Solving one-step and two-step equations – Year 7 maths lesson to download as PDF

Example 1

Solve \(x+8=15\).

Subtract 8 from both sides: \(x=7\).

Example 2

Solve \(x-4=9\) and \(5x=30\).

Add 4: \(x=13\). Divide by 5: \(x=6\).

Example 3

Solve \(\dfrac{x}{3}=7\).

Multiply both sides by 3: \(x=21\).

Example 4

Solve \(3x+5=26\).

Subtract 5: \(3x=21\). Divide by 3: \(x=7\). Check: \(3\times7+5=26\). Correct.

Example 5

Solve \(\dfrac{x}{4}-2=3\).

Add 2: \(\dfrac{x}{4}=5\). Multiply by 4: \(x=20\).

Example 6

Solve \(2x+9=3\).

Subtract 9: \(2x=-6\). Divide by 2: \(x=-3\). Check: \(2\times(-3)+9=3\).

Example 7

Solve \(4(x-3)=20\).

Divide both sides by 4: \(x-3=5\), so \(x=8\). Or expand: \(4x-12=20\), \(4x=32\), \(x=8\).

Example 8

Solve \(5x-3=2x+12\).

Subtract \(2x\): \(3x-3=12\). Add 3: \(3x=15\). Divide: \(x=5\).

Example 9

I think of a number, multiply it by 3 and add 4. The answer is 19. What is my number?

\(3x+4=19\), so \(3x=15\) and \(x=5\).

Example 10

The perimeter of a rectangle with sides \(x\) and \(x+3\) is 26 cm. Find \(x\).

\(2x+2(x+3)=26\), so \(4x+6=26\), \(4x=20\) and \(x=5\).

Example 11

Solve \(4x=10\) and \(3x+2=6\).

\(x=\dfrac{10}{4}=\dfrac{5}{2}\) or 2.5. For the second, \(3x=4\), so \(x=\dfrac{4}{3}\) or \(1\dfrac{1}{3}\).

Example 12

Ben is 5 years older than Cara. Their ages add up to 35. How old is Cara?

Let Cara be \(c\). Ben is \(c+5\). \(c+c+5=35\), so \(2c=30\) and \(c=15\). Cara is 15 and Ben is 20.

Example 13

Solve \(7-2x=1\).

Subtract 7: \(-2x=-6\). Divide by \(-2\): \(x=3\). Check: \(7-2\times3=1\).

Example 14

Solve \(3(2x+1)=2(x+9)\).

Expand: \(6x+3=2x+18\). Subtract \(2x\): \(4x+3=18\). Subtract 3: \(4x=15\). So \(x=3.75\).

Worked example in context

A plumber charges £40 call-out plus £25 per hour. A bill is £140. How many hours did the job take?

Equation: \(40+25h=140\). Subtract 40: \(25h=100\). Divide by 25: \(h=4\). The job took 4 hours.

Example 1

Solve \(4(x-3)=2x+8\).

First we expand the bracket: \(4x-12=2x+8\). There are more x’s on the left, so we subtract \(2x\) from both sides to get \(2x-12=8\). Next we add 12 to both sides, which gives \(2x=20\). Finally we divide both sides by 2, so \(x=10\). Checking: the left side is \(4(10-3)=28\) and the right side is \(2\times10+8=28\). Both sides are equal, so the solution is correct.

Example 2

Solve \(\dfrac{x}{3}+5=9\).

The last thing that happened to x was “add 5”, so we undo it first by subtracting 5 from both sides: \(\dfrac{x}{3}=4\). Now x has been divided by 3, so we multiply both sides by 3 to get \(x=12\). Check: \(12\div3+5=4+5=9\). Undoing the operations in the reverse order is like taking off your shoes before your socks: the last thing done is the first thing undone.

Example 3

The angles in a triangle are \(x\), \(2x\) and \(x+40\) degrees. Find each angle.

The angles of a triangle add up to 180°, so we form the equation \(x+2x+x+40=180\). Collecting like terms gives \(4x+40=180\). Subtracting 40 from both sides gives \(4x=140\), and dividing by 4 gives \(x=35\). The angles are 35°, 70° and 75°. They add up to 180°, which confirms the answer. Many word problems become easy once you have written the right equation.

Example 4

A gym charges a £20 joining fee plus £15 per month. Another gym charges £35 per month and no joining fee. After how many months do the two gyms cost the same?

Let m be the number of months. The first gym costs \(20+15m\) pounds and the second costs \(35m\) pounds. Setting them equal gives \(20+15m=35m\). Subtracting \(15m\) from both sides gives \(20=20m\), so \(m=1\). After one month both gyms cost £35; after that the first gym is cheaper because it charges less each month.