Linear and simultaneous equations. Attnom Academy

M4.8.1 Chapter 1

Harder linear equations

How do I solve 3(x + 2) = 2(x + 5), with a bracket on both sides?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Multiply only the first term in each bracket: 3x + 2 = 2x + 5.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Drop the brackets and keep the numbers: 3 + x + 2 = 2 + x + 5.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Multiply every term in each bracket: 3x + 6 = 2x + 10.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve equations with brackets on both sides
Solve equations with fractions, like (x + 2)/3 = 5

Ravi packs boxes. Each box holds one bag of beans and 2 weights of 1 kg. Every bag weighs the same. Call that weight x kg.

Three boxes weigh 3(x + 2) kg. The bracket tells you that the 3 multiplies everything inside it. You get 3 bags and 3 × 2 = 6 weights.

Draw it first

Open the bracket

1 Draw a rectangle 3 high. Make it x + 2 wide: one piece for the bag and one for the 2 weights.
2 Each of the 3 rows has an x and a 2. Write 3x for the bags and 6 for the weights.
3 The whole rectangle is the same as its two cells: 3(x + 2) = 3x + 6.

To expand a bracket is to multiply every term inside it by the number outside. A term is one part of an expression, with its sign. In (x + 2) the terms are x and +2.

Put 3 of these boxes on one pan of a balance. On the other pan put 2 boxes that each hold one bag and 5 weights. The pans are level, so 3(x + 2) = 2(x + 5). Open the brackets and look at what is on each pan.

After the brackets are opened: 3x + 6 balances 2x + 10. Each small square is a 1 kg weight.

A balance stays level if you take the same from both pans. Take 2 bags off each pan. The left pan keeps 1 bag and 6 weights. The right pan keeps 10 weights.

Take 2x from both sides: x + 6 balances 10. Each small square is a 1 kg weight.

Worked example

Solve 3(x + 2) = 2(x + 5).

Expand both brackets: 3x + 6 = 2x + 10.
Take 2x from both sides: x + 6 = 10.
Take 6 from both sides: x = 4.
Check: the left side is 3(4 + 2) = 18 and the right side is 2(4 + 5) = 18.

Answer: x = 4

Take care with a minus sign inside a bracket. In 4(x − 1) the 4 multiplies the −1 too, so you get 4x − 4. The minus sign stays with the number.

Worked example

Solve 4(x − 1) = 2(x + 5).

Expand both brackets: 4x − 4 = 2x + 10.
Take 2x from both sides: 2x − 4 = 10.
Add 4 to both sides: 2x = 14.
Divide both sides by 2: x = 7.
Check: the left side is 4(7 − 1) = 24 and the right side is 2(7 + 5) = 24.

Answer: x = 7

Now fractions. The equation (x + 2)/3 = 5 says: x + 2, shared into 3 equal parts, gives 5 in each part.

Three parts of 5 make x + 2.

So x + 2 must be 3 × 5 = 15. In symbols, multiply both sides of the equation by 3, the number below the line. Every term on a side is multiplied.

Worked example

Solve (x + 2)/3 = 5.

Multiply both sides by 3. On the left, 3 × (x + 2)/3 is x + 2.
On the right, 3 × 5 = 15. So x + 2 = 15.
Take 2 from both sides: x = 13.
Check: (13 + 2)/3 = 15/3 = 5.

Answer: x = 13

See it three ways

With things

Two boxes each hold one bag and 3 weights of 1 kg. Together they weigh 14 kg. Share 14 kg into 2 equal boxes.

As a picture

In words

Each box weighs 14 ÷ 2 = 7 kg. A box holds a bag and 3 kg, so the bag weighs 7 − 3 = 4 kg.

In symbols

2(x + 3) = 14. Then x + 3 = 7, so x = 4. Or open the bracket first: 2x + 6 = 14, then 2x = 8, so x = 4.

Try it

Test an answer in both sides

You need: paper, a pencil.

Write 3(x + 2) = 2(x + 5).
Put x = 4 in the left side, then in the right side. Write both answers.
Try x = 5 and x = 3 the same way.

What you will see: Only x = 4 gives the same number on both sides. Any other value leaves one side bigger.

If you cannot try it, this is what happens.

A wrong idea many people have

4(x − 1) = 4x − 1.

Why it feels right: The 4 sits next to the x, so it feels as if it only touches the x.

The test that breaks it: Put x = 3. Then 4(3 − 1) = 4 × 2 = 8. But 4x − 1 = 12 − 1 = 11. The two do not match, and 4x − 4 = 8 does.

What is true: The number outside multiplies every term inside the bracket, including the one with the minus sign.

A wrong idea many people have

To clear the 3 in (x + 2)/3 = 5, multiply only the x by 3. That gives 3x + 2 = 5.

Why it feels right: The x is the letter you want, so it feels like the only thing to change.

The test that breaks it: From 3x + 2 = 5 you get x = 1. Put x = 1 in the original: (1 + 2)/3 = 1, not 5.

What is true: Multiply the whole of each side by 3. The whole x + 2 is multiplied, not just the x.

Key words

expand: to remove a bracket by multiplying every term inside it by the number outside.
term: one part of an expression, with its sign, such as 3x or −4.

Back to your guess

The answer is C: Multiply every term in each bracket: 3x + 6 = 2x + 10.

The number outside a bracket multiplies every term inside it. So 3(x + 2) is 3x + 6, and 2(x + 5) is 2x + 10. Then you solve 3x + 6 = 2x + 10, and x = 4.

Keep it

Copy this into your exercise book.

Open a bracket by multiplying every term inside it. Clear a fraction by multiplying both sides by the number below it. Do the same to both sides.

Draw: Three balances in a row: 3x + 6 against 2x + 10, then x + 6 against 10, then x against 4.

M4.8.2 Chapter 2

Changing the subject

The formula v = u + at gives speed. How do I get t on its own?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Divide by a first, then add u: t = v/a + u.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Take u from both sides, then divide everything by a: t = (v − u)/a.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Take away u and a: t = v − u − a.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Rearrange v = u + at to make t the subject
Rearrange a formula where the subject appears twice

Tariq rides a bike. The formula v = u + at gives his speed after a time t. Here u is his speed at the start, and a is how much faster he goes each second. The letter v is his speed at the end. Speeds are in metres per second, and time is in seconds.

In v = u + at, the letter v stands alone on the left side. We call it the subject of the formula. What if you know v, u and a, and want t? Then you must rearrange the formula to make t the subject.

Draw it first

Undo the machine

1 Start with t. Multiply by a to get at. Then add u to get v.
2 To go back, start at v. Undo the adding by taking u away. Then undo the multiplying by dividing by a.
3 The way back is t = (v − u)/a. The whole of v − u is divided by a.

The undo row goes in reverse order, and each of its steps is the opposite of one step in the forward row. Adding and taking away undo each other. Multiplying and dividing undo each other.

A balance shows why. Take u = 2 and a = 3, so v = 2 + 3t. Say t = 4. Then v = 14. The left pan holds 14 weights. The right pan holds 3 bags of t and 2 weights.

v = 14 balances u + at = 2 + 3t. Each small square is one unit.

Take u = 2 weights off both pans. That is the same as taking u from both sides. The left pan keeps 12 weights. The right pan keeps the 3 bags.

v − u = 12 balances at = 3t. Each small square is one unit.

Now share the 12 weights between the 3 bags. Each bag is 4. That is the same as dividing both sides by a. You get t = 12 ÷ 3 = 4.

Worked example

Make t the subject of v = u + at.

Start with v = u + at. The letter t is only in the term at.
Take u from both sides: v − u = at.
Divide both sides by a: (v − u)/a = t.
Write t on the left: t = (v − u)/a. This needs a not to be zero, because you cannot divide by zero.
Check with u = 2, a = 3 and t = 4. Then v = 14, and (14 − 2)/3 = 12/3 = 4.

Answer: t = (v − u)/a. With v = 14, u = 2 and a = 3, t = 4.

The same steps work for any formula. The temperature formula F = 1.8C + 32 turns degrees Celsius into degrees Fahrenheit. To go the other way, make C the subject.

Worked example

Make C the subject of F = 1.8C + 32.

The formula multiplies C by 1.8, then adds 32. Undo the adding first.
Take 32 from both sides: F − 32 = 1.8C.
Divide both sides by 1.8: C = (F − 32)/1.8.
Test it. At normal air pressure, water boils at 212 °F, so F − 32 = 180.
Then C = 180 ÷ 1.8 = 100. Water boils at 100 °C at normal air pressure.

Answer: C = (F − 32)/1.8. For F = 212, C = 100.

Sometimes the subject appears twice. In P = ax + bx the letter x is in both terms. You cannot undo one without the other. Instead, factorise: take x out as a common factor.

ax + bx is a rectangle x high and a + b wide, so ax + bx = x(a + b).

Now x is in one place only. P = x(a + b). Divide both sides by (a + b) and you get x = P/(a + b). This needs a + b not to be zero.

Worked example

Make x the subject of ax + b = cx + d.

Collect the x terms on the left and the other terms on the right. Take cx and b from both sides: ax − cx = d − b.
Factorise the left side. Both terms have x: ax − cx = x(a − c).
Now x is in one place: x(a − c) = d − b.
Divide both sides by (a − c): x = (d − b)/(a − c). This needs a − c not to be zero.
Try a = 5, b = 2, c = 3 and d = 8. Then d − b = 8 − 2 = 6.
And a − c = 5 − 3 = 2.
So x = 6/2 = 3. Put it back: 5 × 3 + 2 = 17, and 3 × 3 + 8 = 17.

Answer: x = (d − b)/(a − c). With a = 5, b = 2, c = 3 and d = 8, x = 3.

See it three ways

With things

Joseph starts a run at 5 metres per second and speeds up by 2 metres per second every second. His speed is now 17 metres per second. How many seconds has he been speeding up?

As a picture

In words

Take 5 away from 17 to get 12. The speed gains 2 every second, so divide 12 by 2. That gives 6 seconds.

In symbols

17 = 5 + 2t, so 2t = 12 and t = 6. Or use the formula: t = (v − u)/a = (17 − 5)/2 = 6.

Try it

Undo a number trick

You need: a friend, paper and a pencil.

Ask your friend to think of a number and keep it secret.
Ask them to multiply it by 3, then add 2, and tell you the answer.
Take away 2, then divide by 3. Tell them their number.

What you will see: You get their number every time. You undid the steps in the opposite order, and each step was undone by its opposite.

If you cannot try it, this is what happens.

A wrong idea many people have

To get t alone in v = u + at, divide v by a and add u. So t = v/a + u.

Why it feels right: The letter a sits next to t, so it feels as if you can deal with it first.

The test that breaks it: Put u = 2, a = 3 and t = 4, so v = 14. Then v/a + u = 14/3 + 2, which is not 4. But (v − u)/a = 12/3 = 4.

What is true: Undo in reverse order. The u was added last, so take it away first. Then divide the whole of v − u by a.

A wrong idea many people have

You can always divide both sides by a letter.

Why it feels right: Dividing by a number such as 3 always works, so a letter feels the same.

The test that breaks it: Put a = 0 in v = u + at. Then v = u + 0 × t, so v = u whatever t is. Knowing v and u tells you nothing about t.

What is true: You can divide by a letter only when it is not zero. Say so when you need it, as in t = (v − u)/a for a not zero.

Key words

subject: the letter that stands alone on one side of a formula, such as v in v = u + at.
rearrange: to change a formula so that a different letter is the subject.
factorise: to write a sum as a product by taking out a common factor, as ax + bx = x(a + b).

Back to your guess

The answer is B: Take u from both sides, then divide everything by a: t = (v − u)/a.

The formula multiplies t by a, then adds u. To undo it, take away u first, then divide the whole of v − u by a. So t = (v − u)/a.

Keep it

Copy this into your exercise book.

Undo the steps in reverse order, doing the same to both sides. If the subject is in two terms, collect them and factorise. Never divide by zero.

Draw: A machine with two rows: t, × a, at, + u, v, and the row back: v, − u, at, ÷ a, t.

M4.8.3 Chapter 3

Inequalities

The answer to 3x − 4 < 11 is not one number. How do I show every answer?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A x = 5, because 3 × 5 − 4 = 11.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x > 5, because the sign flips when you divide by 3.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x < 5, and every number below 5 fits.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve 3x − 4 < 11 and show it on a number line
Solve an inequality where you must flip the sign
List the whole numbers that fit −2 < x ≤ 3

A bridge sign says: loads must be less than 5 tonnes. A load of 4 tonnes is allowed. So is 3, and so is 0.5. Many numbers fit. We write the rule as x < 5.

The sign < means "is less than". The sign > means "is more than". The sign ≤ means "is less than or equal to". The sign ≥ means "is more than or equal to". A statement with one of these signs is an inequality.

Draw it first

Show x < 5 on a number line

1 Draw a number line. Put an open circle at 5. It is open because 5 is not allowed: x is less than 5, not equal to 5.
2 Draw a thick bar from the circle towards the smaller numbers. The arrow says it goes on for ever.
3 Dots on the whole numbers 4, 3, 2, 1, 0 and −1 show a few of the numbers that fit. The arrow shows there are more.

An open circle means the end number is not included. It goes with < and >. A filled circle means the end number is included. It goes with ≤ and ≥.

Now solve 3x − 4 < 11. Treat it like an equation and do the same to both sides. Adding 4 to both sides gives 3x < 15. A balance shows this. Say a bag weighs 4 kg. Then 3 bags weigh 12 kg, and the other pan holds 15 kg. The 15 kg side is heavier, so the beam tilts.

With x = 4, 3x is less than 15. The beam tilts, and it stays tilted if you take the same from both sides. Each small square is 1 kg.

Worked example

Solve 3x − 4 < 11 and show it on a number line.

Add 4 to both sides. The sign stays the same: 3x < 15.
Divide both sides by 3. The sign stays the same: x < 5.
Test it. When x = 4, 3x − 4 = 8, which is less than 11. When x = 6, 3x − 4 = 14, which is not.
On the number line put an open circle at 5 with a bar going left.

Answer: x < 5

x = 5 gives exactly 11, which is not less than 11. So 5 is left out and the circle is open.

Now a trap. On a number line the bigger number is always further right. 2 is left of 5, so 2 < 5. Multiply both numbers by −1 and look again.

2 < 5, but after multiplying by −1 the order turns round: −2 > −5.

Multiplying by −1 swaps left and right. So multiplying or dividing both sides by a negative number must flip the sign: < becomes > and > becomes <. Adding or taking away never flips it. Neither does multiplying or dividing by a positive number.

Worked example

Solve 7 − 2x > 1.

Take 7 from both sides. The sign stays the same: −2x > −6.
Divide both sides by −2. The number is negative, so the sign flips: x < 3.
Test it. When x = 0, 7 − 2x = 7, which is more than 1. When x = 4, 7 − 2x = −1, which is not.

Answer: x < 3

x < 3: the open circle leaves 3 out.

Sometimes x is boxed in on both sides. The inequality −2 < x ≤ 3 means x is more than −2 and also at most 3. The circle at −2 is open, and the circle at 3 is filled.

Here −2 < x ≤ 3, and the whole numbers that fit are the five dots.

Worked example

List the whole numbers that fit −2 < x ≤ 3.

Read it as two rules. x is more than −2, and x is at most 3.
The circle at −2 is open, so −2 does not fit. The first whole number is −1.
The circle at 3 is filled, so 3 does fit. The last whole number is 3.
The whole numbers are −1, 0, 1, 2, 3.
Count them. There are 5 whole numbers.

Answer: −1, 0, 1, 2, 3

See it three ways

With things

A shelf holds fewer than 5 books. It might hold none. How many books could be on it?

As a picture

In words

The shelf holds 0, 1, 2, 3 or 4 books. It cannot hold 5, because 5 is not fewer than 5.

In symbols

x is a whole number with 0 ≤ x < 5.

Try it

Mark a rule with stones

You need: paper and a pencil, a few small stones or coins.

Draw a number line from 0 to 8 and write the numbers under it.
The rule is x ≥ 3. Put a stone on every whole number that fits.
Mark 3 with a filled circle and draw a bar with an arrow to the right.

What you will see: The stones sit on 3, 4, 5, 6, 7 and 8. The number 3 has a stone because ≥ includes it, so the circle is filled. The arrow shows that 9, 10 and every bigger number fit too.

If you cannot try it, this is what happens.

A wrong idea many people have

You flip the sign whenever you divide.

Why it feels right: Flipping is the new rule, so it feels as if it applies to every division.

The test that breaks it: Solve 3x < 15. Dividing by 3 gives x < 5. Put x = 4: 3 × 4 = 12, which is less than 15. A flipped sign, x > 5, would say 4 is not allowed.

What is true: The sign flips only when you multiply or divide both sides by a negative number. Dividing by 3 is dividing by a positive number.

A wrong idea many people have

x > 3 and x ≥ 3 mean the same.

Why it feels right: The two signs look almost the same, and both point to the bigger numbers.

The test that breaks it: Put x = 3. Then 3 ≥ 3 is true, because 3 equals 3. But 3 > 3 is false.

What is true: x > 3 leaves 3 out, with an open circle. x ≥ 3 keeps 3, with a filled circle.

Key words

inequality: a statement that one thing is less than, more than, at most or at least another. Two examples are x < 5 and x ≥ 3.
open circle: a ring on a number line. It shows the end number is not included.
filled circle: a solid dot on a number line. It shows the end number is included.

Back to your guess

The answer is C: x < 5, and every number below 5 fits.

An inequality has many answers. 3x − 4 < 11 gives 3x < 15, so x < 5. Every number below 5 fits. The sign flips only when you multiply or divide both sides by a negative number.

Keep it

Copy this into your exercise book.

Open circle: not included. Filled circle: included. Solve like an equation. Flip the sign only when you multiply or divide by a negative number.

Draw: A number line with an open circle at 5 and a bar going left, and the words x < 5.

M4.8.4 Chapter 4

Equations of straight lines

A straight line goes through (1, 3) and (4, 9). How do I find its equation?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A y = 6x − 3, because the line goes up 6.
You picked A. Read on. At the end, the chapter comes back to your guess.
B y = 0.5x + 2.5, because it goes across 3 for every 6 up.
You picked B. Read on. At the end, the chapter comes back to your guess.
C y = 2x + 1, because it goes up 2 for every 1 across.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find the equation of a line through two points
Spot parallel and perpendicular lines from their gradients

A path goes from the point A (1, 3) to the point B (4, 9). Going across, it climbs. How steep is it? Count the steps across, and count how far it goes up.

Draw it first

Rise over run

1 Draw the line through A and B.
2 From A go across to the right until you are below B. This is the run: 4 − 1 = 3.
3 Then go up to B. This is the rise: 9 − 3 = 6.

The gradient is how steep the line is. It is the rise divided by the run: 6 ÷ 3 = 2. For the points (x₁, y₁) and (x₂, y₂), the gradient is m = (y₂ − y₁)/(x₂ − x₁). Take the points in the same order on the top and the bottom.

The equation of a straight line is y = mx + c. The number m is the gradient. The number c is where the line crosses the y axis. For our line m = 2, so y = 2x + c. To find c, put in the coordinates of any point on the line.

The line crosses the y axis at 1, so c = 1.

Worked example

Find the equation of the line through (1, 3) and (4, 9).

Gradient: m = (9 − 3)/(4 − 1) = 6/3 = 2.
The line is y = 2x + c.
Put in the point (1, 3): 3 = 2 × 1 + c, so c = 1.
The equation is y = 2x + 1.
Check with the other point (4, 9): 2 × 4 + 1 = 9.

Answer: y = 2x + 1

A line that goes down as you move right has a negative rise. Its gradient is negative. From J (2, 7) to K (5, 1) the run is 3 and the rise is −6, so the gradient is −2.

A line that falls to the right has a negative gradient.

Worked example

Find the equation of the line through (2, 7) and (5, 1).

Rise: 1 − 7 = −6. Run: 5 − 2 = 3.
Gradient: m = −6/3 = −2. The line falls, so the gradient is negative.
The line is y = −2x + c. Put in the point (2, 7): 7 = −2 × 2 + c, so c = 11.
The equation is y = −2x + 11.
Check with (5, 1): −2 × 5 + 11 = 1.

Answer: y = −2x + 11

Two lines are parallel when they never meet. Parallel lines go up at the same rate, so they have the same gradient. Lines with the same gradient and different values of c are parallel.

The same gradient, 2, makes the lines parallel.

Two lines are perpendicular when they cross at a right angle. Their gradients multiply to give −1. To see why, take a step "across 2, up 1" along a line through the origin. A quarter turn about the origin sends it to "across −1, up 2". The first line has gradient 1/2 and the turned line has gradient −2.

A is (2, 1). A quarter turn takes it to B, which is (−1, 2). The gradients 0.5 and −2 multiply to −1.

In general, a step "across a, up b" turns into "across −b, up a". The gradients are b/a and −a/b, and their product is −1. This works when neither line is upright. An upright line has no gradient, and it is perpendicular to a flat line.

The gradients are 2 and −0.5. Their product is −1, so the lines are perpendicular.

The line y = 0.5x + 3 goes through T (4, 5). It meets y = −2x + 5 at a right angle.

Worked example

A line has the equation y = −2x + 5. Find the line perpendicular to it through (4, 5).

The first line has gradient −2. For perpendicular lines the gradients multiply to give −1.
So −2 × m = −1, which gives m = 0.5.
The new line is y = 0.5x + c. It passes through (4, 5).
Put in x = 4 and y = 5: 5 = 0.5 × 4 + c, so 5 = 2 + c.
So c = 5 − 2 = 3, and the line is y = 0.5x + 3.
Check: the point (4, 5) gives 0.5 × 4 + 3 = 5. And −2 × 0.5 = −1.

Answer: y = 0.5x + 3

See it three ways

With things

A ramp starts 2 metres up a wall and reaches 8 metres at 2 metres across. How steep is it, and what is its rule?

As a picture

In words

The line starts at 2 on the y axis. For every 2 across it goes up 6, so for every 1 across it goes up 3.

In symbols

m = (8 − 2)/(2 − 0) = 3 and c = 2, so y = 3x + 2.

Try it

Find a line from two points

You need: squared paper, a ruler, a pencil.

Plot the points (2, 2) and (6, 5) and draw the line through them.
Mark the run from the first point across to the right. Count it: 4.
Mark the rise up to the second point. Count it: 3. Write the gradient, 3 ÷ 4.

What you will see: The gradient is 0.75. Where the line meets the y axis you can read c = 0.5, so the equation is y = 0.75x + 0.5.

If you cannot try it, this is what happens.

A wrong idea many people have

A line perpendicular to a line with gradient 2 has gradient −2.

Why it feels right: Perpendicular lines point in opposite directions, so the opposite number feels right.

The test that breaks it: Take y = 2x and y = −2x. Their gradients multiply to 2 × (−2) = −4, not −1. Drawn on squared paper, the two lines do not meet at a right angle. The line y = −0.5x does.

What is true: Perpendicular gradients multiply to −1. Turn the number upside down and change its sign: 2 becomes −1/2, which is −0.5.

A wrong idea many people have

The gradient is the run divided by the rise.

Why it feels right: Both numbers are on the picture, so the order feels like a choice.

The test that breaks it: The line y = 2x goes up 2 for every 1 across. Run ÷ rise gives 1 ÷ 2 = 0.5, but the steep line has gradient 2.

What is true: The gradient is the rise divided by the run. A steeper line has a bigger gradient.

Key words

gradient: how steep a line is: the rise divided by the run, m = (y₂ − y₁)/(x₂ − x₁).
parallel: lines that never meet. They have the same gradient.
perpendicular: lines that cross at a right angle. Their gradients multiply to give −1.

Back to your guess

The answer is C: y = 2x + 1, because it goes up 2 for every 1 across.

The line rises 6 while it runs 3 across, so the gradient is 6 ÷ 3 = 2. Putting the point (1, 3) into y = 2x + c gives c = 1. The equation is y = 2x + 1.

Keep it

Copy this into your exercise book.

Gradient m = rise ÷ run. Put one point in y = mx + c to find c. Parallel: same m. Perpendicular: the m values multiply to −1.

Draw: A line through two points with a dashed run and rise, and y = 2x + 1 written beside it.

M4.8.5 Chapter 5

Two equations: adding or taking away

I know x + y = 10 and x − y = 2. How do I find both numbers?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A x = 5 and y = 5, because 5 + 5 = 10.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x = 6 and y = 4, because 6 + 4 = 10 and 6 − 4 = 2.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x = 8 and y = 2, because 8 + 2 = 10.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve two equations by getting rid of one letter
Multiply an equation first so the numbers match

Amina thinks of two numbers, x and y. She says: "Their sum is 10." That alone has many answers. 4 and 6 work, and so do 5 and 5. One clue is not enough for two unknown numbers.

Every pair adds to 10. Only x = 6 and y = 4 also gives x − y = 2.

Amina adds a second clue: "Their difference is 2", so x − y = 2. Now both equations must be true at the same time. Two equations like this are called simultaneous equations.

Look at the y terms. One equation has +y and the other has −y. If you add the two equations, the y terms cancel. This is elimination: you remove one letter.

Why may you add two equations? x + y and 10 are the same amount, and so are x − y and 2. Add equal amounts to equal amounts and the totals are equal too. It is like putting the same weight on both pans of a level balance. Taking equal amounts from equal amounts works the same way.

Draw it first

Add the equations

1 Write the two equations one under the other, with the same letters in the same columns.
2 The y terms are +y and −y. They cancel when you add, so mark them.
3 Add the left sides and add the right sides. x + x = 2x, and 10 + 2 = 12. So 2x = 12.

Worked example

Solve x + y = 10 and x − y = 2.

Add the equations. The y terms cancel: 2x = 12.
Divide both sides by 2: x = 6.
Put x = 6 in x + y = 10. Then 6 + y = 10, so y = 4.
Check in the other equation, x − y = 2: 6 − 4 = 2.

Answer: x = 6, y = 4

Sometimes both equations have the same term. Then the sign of the terms is the same, so you subtract. Take 3x + 2y = 19 and x + 2y = 9. Both have +2y.

The 2y terms match and have the same sign, so subtract: 3x − x = 2x and 19 − 9 = 10.

Worked example

Solve 3x + 2y = 19 and x + 2y = 9.

The y terms are both +2y, so subtract equation (2) from equation (1). Then 2x = 10.
Divide both sides by 2: x = 5.
Put x = 5 in equation (2): 5 + 2y = 9, so 2y = 4 and y = 2.
Check in equation (1): 3 × 5 + 2 × 2 = 19.

Answer: x = 5, y = 2

The rule is: if the signs of the matching terms are different, add. If they are the same, subtract. Always do the same to every term, on both sides of the equals sign.

What if no letter matches? Multiply an equation first. Multiplying every term of an equation by the same number keeps it true. Use it to make one letter match.

Multiply (1) by 3 and (2) by 2. Then both have 6x.

Call the new equations (3) and (4). The 6x terms match and have the same sign. Subtract: 9y − 4y = 5y and 36 − 26 = 10.

Worked example

Solve 2x + 3y = 12 and 3x + 2y = 13.

Make the x terms match. Multiply (1) by 3 and (2) by 2. Call the new equations (3) and (4): 6x + 9y = 36 and 6x + 4y = 26.
The x terms have the same sign, so subtract (4) from (3): 5y = 10.
Divide both sides by 5: y = 2.
Put y = 2 in equation (1): 2x + 6 = 12, so 2x = 6 and x = 3.
Check in equation (2): 3 × 3 + 2 × 2 = 13.

Answer: x = 3, y = 2

See it three ways

With things

Two bags of rice and one bag of beans weigh 7 kg. One bag of rice and one bag of beans weigh 4 kg. How heavy is a bag of rice?

As a picture

In words

The first weighing has one more bag of rice. The extra 7 − 4 = 3 kg is one bag of rice. The beans are 4 − 3 = 1 kg.

In symbols

2r + b = 7 and r + b = 4. Subtract: r = 3. Put it back: 3 + b = 4, so b = 1.

Try it

The sum and the difference

You need: paper and a pencil, a friend.

Think of two whole numbers. Say them to yourself and keep them secret.
Tell your friend their sum and their difference, the bigger minus the smaller.
Your friend adds the sum and the difference, then halves the answer. That is your bigger number.

What you will see: The sum plus the difference is twice the bigger number, because the smaller number cancels. Take the bigger number from the sum to find the smaller.

If you cannot try it, this is what happens.

A wrong idea many people have

To eliminate a letter you always add the equations.

Why it feels right: The first example was solved by adding, so adding feels like the method.

The test that breaks it: Take x + y = 10 and x + 3y = 14. Adding gives 2x + 4y = 24, which still has both letters. Subtracting gives −2y = −4, so y = 2, which has one letter.

What is true: Add when the matching terms have different signs. Subtract when they have the same sign.

A wrong idea many people have

When you multiply an equation, you multiply only the term you want to match.

Why it feels right: You are only trying to change one number, so changing one number feels enough.

The test that breaks it: Multiply x + y = 4 by 3 on the x only: 3x + y = 4. Now x = 1, y = 3 does not fit it: 3 + 3 = 6, not 4. But 1 + 3 = 4 fitted the original.

What is true: Multiply every term on both sides by the same number. Then the new equation says the same thing as the old one.

Key words

simultaneous equations: two or more equations that must be true at the same time.
elimination: removing one letter by adding or subtracting the equations.

Back to your guess

The answer is B: x = 6 and y = 4, because 6 + 4 = 10 and 6 − 4 = 2.

Both equations must be true at once. x = 6 and y = 4 fit both: 6 + 4 = 10 and 6 − 4 = 2. The other pairs fit only the first equation.

Keep it

Copy this into your exercise book.

Match one letter, multiplying an equation if you must. Different signs: add. Same signs: subtract. Find the other letter, then check in both equations.

Draw: Two equations under each other with the matching column hatched, a line, and the result row under it.

M4.8.6 Chapter 6

Two equations: substitution

One equation says y = 2x + 1. How do I use that in 3x + y = 11?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Put 2x in place of y: 3x + 2x = 11.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Put 2x + 1 in place of y: 3x + (2x + 1) = 11.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Try values of x until 3x + y = 11 works.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve two equations by putting one into the other
Check an answer in both equations

Leila has two clues about two numbers x and y. Clue one: y = 2x + 1. Clue two: 3x + y = 11. The first clue already says what y is, in terms of x.

To substitute is to put one thing in place of another. If y is the same as 2x + 1, then wherever y appears you can write 2x + 1 instead. Put it in brackets, so it stays one whole thing.

Draw it first

Put it in

1 Start with the second equation, 3x + y = 11.
2 The first equation says y is 2x + 1. Put 2x + 1 in brackets in place of y.
3 Now only x is left. Collect the x terms: 3x + 2x = 5x.
4 Take 1 from both sides to get 5x = 10. Then x = 2.

Once x = 2, find y from the first equation: y = 2 × 2 + 1 = 5. The answer is a pair: x = 2 and y = 5.

Worked example

Solve y = 2x + 1 and 3x + y = 11.

Put 2x + 1 in place of y in 3x + y = 11: 3x + (2x + 1) = 11.
Collect the x terms: 5x + 1 = 11, so 5x = 10.
Divide both sides by 5: x = 2.
Find y from y = 2x + 1: y = 2 × 2 + 1 = 5.
Check in both equations. In y = 2x + 1: 2 × 2 + 1 = 5. In 3x + y = 11: 3 × 2 + 5 = 11.

Answer: x = 2, y = 5

Why check in both equations? A wrong pair can fit one equation and miss the other. The pair x = 3, y = 7 fits the first, but not the second.

(1) is y = 2x + 1 and (2) is 3x + y = 11. Only the first pair fits both.

Often neither equation gives a letter on its own. Then first make a letter the subject of one equation. This is the skill from the second chapter of this book.

The y had a minus sign in front, so add y to both sides first. Now y = 3x − 5.

Worked example

Solve x + 2y = 11 and 3x − y = 5.

Make y the subject of the second equation: y = 3x − 5.
Put 3x − 5 in place of y in the first: x + 2(3x − 5) = 11.
Expand the bracket: x + 6x − 10 = 11, so 7x = 21.
Divide both sides by 7: x = 3.
Find y: y = 3 × 3 − 5 = 4.
Check in the first equation: 3 + 2 × 4 = 11. In the second: 3 × 3 − 4 = 5.

Answer: x = 3, y = 4

Worked example

Solve x = y + 4 and 2x + 3y = 18.

Put y + 4 in place of x in the second equation: 2(y + 4) + 3y = 18.
Expand and collect: 2y + 8 + 3y = 18, so 5y + 8 = 18 and 5y = 10.
Divide both sides by 5: y = 2.
Find x from x = y + 4: x = 2 + 4 = 6.
Check in both equations. In x = y + 4: 6 = 2 + 4. In 2x + 3y = 18: 2 × 6 + 3 × 2 = 18.

Answer: x = 6, y = 2

See it three ways

With things

Plot both rules on squared paper. First write 3x + y = 11 as y = −3x + 11 by taking 3x from both sides. Look for the one point that is on both lines.

As a picture

In words

The lines cross at one point. That point is on both lines, so both equations are true there. It is x = 2 and y = 5.

In symbols

3x + (2x + 1) = 11 gives x = 2. Then y = 2 × 2 + 1 = 5.

Bars show a substitution well. Say two numbers add to 13, and x is 3 more than y. The bar for x is the bar for y with 3 added. So x can be replaced by y + 3.

x is y and 3 more, so x = y + 3.

Try it

A number puzzle

You need: paper and a pencil.

Two numbers add to 13. One is 3 more than the other. Write x = y + 3 and x + y = 13.
Put y + 3 in place of x in the second equation. Solve it for y.
Find x, then check both clues with the two numbers.

What you will see: You get y = 5 and x = 8. The numbers 8 and 5 add to 13, and 8 is 3 more than 5.

If you cannot try it, this is what happens.

A wrong idea many people have

To put y = x + 2 into 10 − y, write 10 − x + 2.

Why it feels right: You just write the new letters where y was, so it feels like no more is needed.

The test that breaks it: Take x = 1, so y = 3. Then 10 − y = 7. But 10 − x + 2 = 11. With brackets, 10 − (x + 2) = 7.

What is true: Put the whole expression in brackets. The minus sign then applies to every term inside.

A wrong idea many people have

It is enough to check an answer in one equation.

Why it feels right: You used both equations to find it, so it feels as if both must fit.

The test that breaks it: x = 3 and y = 7 fit y = 2x + 1, because 2 × 3 + 1 = 7. But 3 × 3 + 7 = 16 in the second equation, not 11.

What is true: An answer must fit both equations. Check it in each one.

Key words

substitute: to put one expression in place of a letter that it equals.

Back to your guess

The answer is B: Put 2x + 1 in place of y: 3x + (2x + 1) = 11.

The equation says y is the whole of 2x + 1, so the whole of it goes in place of y. That gives 3x + (2x + 1) = 11, which has only x in it. Then x = 2 and y = 5.

Keep it

Copy this into your exercise book.

Put one equation into the other, in brackets. Solve for one letter, find the other, then check in both equations.

Draw: Two equations, an arrow with the words put in, and the new equation with only one letter.

M4.8.7 Chapter 7

Two equations on a graph

The lines y = x + 1 and y = 7 − x cross at (3, 4). What does that point tell me?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The two lines have the same gradient.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x = 3 and y = 4 make both equations true.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x = 4 and y = 3 make both equations true.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve two equations by finding where their lines cross
Explain why parallel lines give no answer

The equation y = x + 1 has many pairs that fit it. x = 0 and y = 1 fit. So do x = 2 and y = 3. Each pair is a point, and the points form a straight line.

A second equation, y = 7 − x, makes a second line. A pair that fits both equations must be on both lines. Two different lines meet at one point at most.

Draw it first

Draw both lines

1 Draw the line y = x + 1. It crosses the y axis at 1 and rises 1 for each 1 across.
2 Draw the line y = 7 − x. It crosses the y axis at 7 and falls 1 for each 1 across.
3 Mark the point where they cross. Read across and up: x = 3 and y = 4.

The crossing point is the intersection of the lines. Its coordinates are the solution of the two equations: the pair that makes both true.

Worked example

Solve y = x + 1 and y = 7 − x by drawing graphs.

Table for y = x + 1. When x is 0, 3 and 6, y is 1, 4 and 7.
Table for y = 7 − x. When x is 0, 3 and 6, y is 7, 4 and 1.
Plot the points and draw both lines. They cross at (3, 4).
So x = 3 and y = 4.
Check in both equations: 3 + 1 = 4 and 7 − 3 = 4.

Answer: x = 3, y = 4

A graph shows a solution to the accuracy of the drawing. When lines cross between the grid lines, you can only estimate. Take y = 2x and y = 5 − x.

P is between 1 and 2 across, and between 3 and 4 up. The graph gives an estimate.

Algebra gives the exact answer. Put the right sides equal: 2x = 5 − x, so 3x = 5 and x = 5/3, which is about 1.7. Then y = 2x is about 3.3. A graph and algebra agree, but only algebra is exact.

Some lines never cross. y = 2x + 1 and y = 2x + 5 both go up 2 for each 1 across. They stay the same distance apart. Lines with the same gradient and different values of c are parallel.

The lines never cross, so no pair of numbers fits both equations.

Worked example

Solve y = 2x + 1 and y = 2x + 5.

Both lines have gradient 2.
The lines cross the y axis at 1 and at 5. These are different, so the lines are parallel.
Check with algebra. Put the right sides equal: 2x + 1 = 2x + 5. Take 2x from both sides: 1 = 5.
That is false for every x, so there is no solution.

Answer: No solution. Both gradients are 2.

There is one more case. The equations y = 2x + 1 and 2y = 4x + 2 look different. Divide the second by 2 and you get y = 2x + 1 again. The two equations make the same line, so every point on it is a solution. There are infinitely many solutions.

Two equations for one line: infinitely many solutions.

Worked example

Two tanks. Tank A holds 40 litres and gains 5 litres each minute. Tank B holds 10 litres and gains 8 litres each minute. When do they hold the same?

After t minutes tank A holds V = 5t + 40 litres and tank B holds V = 8t + 10 litres.
They hold the same where the lines cross, so 5t + 40 = 8t + 10.
Take 5t and 10 from both sides: 30 = 3t, so t = 10.
Find V from tank A: V = 5 × 10 + 40 = 90 litres.
Check with tank B: V = 8 × 10 + 10 = 90 litres.

Answer: After 10 minutes, with 90 litres in each tank.

Tank B starts lower but fills faster, so the lines cross at P.

See it three ways

With things

Try x = 1, 2, 3 and 4 in both rules. The rule y = 2x gives 2, 4, 6 and 8. The rule y = 9 − x gives 8, 7, 6 and 5.

As a picture

In words

Both rules give 6 when x = 3. The two lines cross at the point where x = 3 and y = 6.

In symbols

2x = 9 − x, so 3x = 9 and x = 3. Then y = 2 × 3 = 6.

Try it

Find where two lines cross

You need: squared paper, a ruler, a pencil.

Make a table for y = 3x and for y = x + 4, with x from 0 to 4.
Plot the points and draw both lines with the ruler.
Read the point where the lines cross, then check it in both rules.

What you will see: The lines cross at x = 2, y = 6. Both rules give 6 when x = 2: 3 × 2 = 6 and 2 + 4 = 6.

If you cannot try it, this is what happens.

A wrong idea many people have

If two lines do not cross at a grid point, there is no solution.

Why it feels right: Whole numbers are quick to read, so a crossing between the grid lines feels like a miss.

The test that breaks it: y = 2x and y = 5 − x cross at x = 5/3, which is about 1.7. Put in x = 5/3: 2 × 5/3 = 10/3, and 5 − 5/3 = 10/3. Both sides are equal.

What is true: Lines with different gradients always cross, at one point. Its coordinates need not be whole numbers.

A wrong idea many people have

Any two lines meet somewhere.

Why it feels right: Lines are drawn as going on for ever, so they feel as if they must meet.

The test that breaks it: y = 2x + 1 and y = 2x + 5 are always 4 apart going up. At x = 0 they are 1 and 5, and at x = 10 they are 21 and 25.

What is true: Lines with the same gradient and different values of c never meet. Two equations like that have no solution.

Key words

intersection: the point where two lines cross.
solution: the pair of numbers that makes both equations true: the coordinates of the intersection.

Back to your guess

The answer is B: x = 3 and y = 4 make both equations true.

A point on a line fits that line's equation. A point on both lines fits both equations. So the crossing point (3, 4) means x = 3 and y = 4: 3 + 1 = 4 and 7 − 3 = 4.

Keep it

Copy this into your exercise book.

The crossing point of two lines is the solution. Parallel lines: no solution. The same line: infinitely many. A graph estimates, algebra is exact.

Draw: Two lines crossing at a point, with the point named and its x and y read off the axes.

M4.8.8 Chapter 8

Word problems and regions

A yard has hens and goats: 30 heads and 84 legs. How many of each?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 18 hens and 12 goats.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 12 hens and 18 goats.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 15 hens and 15 goats.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Write two equations from a word problem and solve them
Shade the region on a graph that fits two inequalities

A word problem gives facts about two unknown numbers. Turn each fact into an equation. The two equations together are simultaneous equations, and you can solve them with the methods of this book.

Take the yard. Let x be the number of hens and y the number of goats. Every animal has one head, so x + y = 30. A hen has 2 legs and a goat has 4, so 2x + 4y = 84.

Draw it first

Facts into equations

1 Count heads. One head each: x + y = 30.
2 Count legs. 2 for each hen and 4 for each goat: 2x + 4y = 84.

Worked example

A yard has hens and goats: 30 heads and 84 legs. Hens have 2 legs and goats have 4. How many of each?

Let x be the number of hens and y the number of goats.
Heads: x + y = 30. Legs: 2x + 4y = 84.
Multiply the first equation by 2: 2x + 2y = 60. Subtract it from the second: 2y = 24.
Divide by 2: y = 12. Then x = 30 − 12 = 18.
Check the legs: 2 × 18 + 4 × 12 = 36 + 48 = 84.

Answer: 18 hens and 12 goats

Always check against the words, not only the equations. 18 + 12 = 30 heads is right, and 84 legs is right. Make sure the answers make sense: a number of animals is a whole number and is not negative.

Now a shape. A rectangle has a perimeter of 34 cm. Its length is 5 cm more than its width. Let l be the length and w the width, in cm.

The perimeter is 2l + 2w. The length l is w + 5.

Worked example

A rectangle has a perimeter of 34 cm. Its length is 5 cm more than its width. Find the sides.

Let l be the length and w the width, both in cm.
The perimeter gives 2l + 2w = 34. The length gives l = w + 5.
Put w + 5 in place of l: 2(w + 5) + 2w = 34, so 4w + 10 = 34.
Take 10 from both sides: 4w = 24, so w = 6.
Then l = 6 + 5 = 11.
Check: 2 × 11 + 2 × 6 = 34 cm.

Answer: The width is 6 cm and the length is 11 cm.

Now inequalities with two letters. The inequality y > x is true for many points, not for one. The points that make it true fill a region of the graph. The line y = x is the boundary.

Draw the boundary solid when the sign is ≤ or ≥, because points on the line fit. Draw it dashed when the sign is < or >, because points on the line do not fit. Then test a point to find which side to hatch. Hatching means shading with lines, so a question that says shade the region wants the same thing.

Draw it first

Hatch a region

1 Draw each boundary. x = 1 and y = 6 are solid. y = x is dashed, because the sign is strict.
2 Choose a point that is on none of the lines, such as T (2, 4), and test it in each inequality.
3 Hatch the region that fits all three. T fits each one: 2 ≥ 1, 4 > 2 and 4 ≤ 6.

To find the side of one line, test a point that is not on it. For y > x + 1, try (0, 0). It gives 0 > 1, which is false, so the origin is on the wrong side. Hatch the other side, above the line.

Worked example

Show the region that fits x ≥ 1, y ≥ 2 and x + y ≤ 8.

Draw the solid lines x = 1, y = 2 and x + y = 8. They are solid because the signs are ≥ and ≤.
Test (0, 0) in x + y ≤ 8: 0 + 0 = 0, and 0 ≤ 8 is true. So hatch the side of x + y = 8 that holds the origin.
Keep only the part with x ≥ 1 and y ≥ 2. The smallest y in the region is 2.
The corners are where pairs of lines meet. Where y = 2 meets x + y = 8, x + 2 = 8, so x = 6.
Where x = 1 meets x + y = 8, 1 + y = 8, so y = 7.
The region is a triangle with corners (1, 2), (6, 2) and (1, 7).

Answer: The triangle with corners (1, 2), (6, 2) and (1, 7).

The hatched triangle fits all three. Its edges are solid, so points on the edges count.

See it three ways

With things

Amina takes at most 6 pieces of fruit: x apples and y mangoes. She wants at least 1 apple and at least 1 mango.

As a picture

In words

Every whole-number point inside the triangle, or on its edge, is a basket Amina can take. For example, 2 apples and 3 mangoes is 5 pieces.

In symbols

x ≥ 1, y ≥ 1 and x + y ≤ 6.

Try it

Hatch your own region

You need: squared paper, a ruler, a pencil.

Draw the lines x = 1 and y = 1 solid, and x + y = 7 dashed.
Test the point (2, 2) in all three signs: x ≥ 1, y ≥ 1 and x + y < 7.
Hatch the triangle that fits all three.

What you will see: The point (2, 2) fits, so it is inside. The dashed edge is not included. So (3, 4), where x + y = 7, is not in the region.

If you cannot try it, this is what happens.

A wrong idea many people have

You can pick either side of the boundary to hatch without testing.

Why it feels right: The sign y > x + 2 sounds like "up", so it feels as if the side is obvious.

The test that breaks it: Test (0, 0) in y > x + 2. It gives 0 > 2, which is false. So the side with the origin is not the region. Test (0, 5): 5 > 2 is true.

What is true: Put a point that is not on the line into the inequality. If it is true, hatch its side. If it is false, hatch the other side.

A wrong idea many people have

A boundary line is always drawn solid.

Why it feels right: Most of the lines in a graph are solid, so it feels like the normal way.

The test that breaks it: Take y < x. The point (3, 3) is on the line y = x. Put it in: 3 < 3 is false. So (3, 3) is not in the region, and the line must show that it is left out.

What is true: A solid line says the points on it are included (≤ or ≥). A dashed line says they are not (< or >).

Key words

region: the part of a graph where every point fits the inequalities.
boundary: the line that edges a region. It is solid when included and dashed when not.

Back to your guess

The answer is A: 18 hens and 12 goats.

Hens have 2 legs and goats have 4. 18 hens and 12 goats give 36 + 48 = 84 legs and 30 heads. The other pairs have 30 heads, but the legs are 96 and 90.

Keep it

Copy this into your exercise book.

Name the unknowns, write a fact as each equation, solve, check in the words. For a region: solid for ≤ and ≥, dashed for < and >, and test a point.

Draw: A triangle made of three lines, hatched inside, with one test point marked.

Open the bracket

Test an answer in both sides

Undo the machine

Undo a number trick

Show x < 5 on a number line

Mark a rule with stones

Rise over run

Find a line from two points

Add the equations

The sum and the difference

Put it in

A number puzzle

Draw both lines

Find where two lines cross

Facts into equations

Hatch a region

Hatch your own region

Key words, A to Z