Solving harder equations

Get the x terms on one side and the numbers on the other, doing the same to both sides; expand any brackets first. Then check by substituting into the first equation.

Two phone plans: 7 dollars plus 3 a gigabyte, or 5 a gigabyte minus a 9 dollar credit. When do they cost the same? 3x + 7 = 5x − 9.

In Year 7 you solved equations with the balance method, even with $x$ on both sides. Year 8 equations add negatives, brackets and fractions, where blocks on a scale no longer help. The rule is the same: do the same to both sides, until $x$ is on its own.

Then check: put your answer back into the first equation. If both sides are equal, it is right.

Predict: Read the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
Try: Try it on the board and compare the result with your prediction.
Investigate: Work out why it happens, one step at a time.
Modify: Change the problem on purpose and see what happens.
Practise: A few on your own, to check you have it.
Make: Make your own problem using what you have learned. Expect a few tries; that is normal.

Explain it to me

Same to both sides, then check · Sound on for the voiceover

To solve an equation, do the same to both sides until $x$ is on its own:

Expand any brackets.
Take the smaller $x$ term from both sides, so $x$ is on one side only.
Undo the numbers with the inverse operation (add to undo $-$ , subtract to undo $+$ ).
Divide (or multiply) to get $x$ .

$3x + 7 = 5x - 9 \to 7 = 2x - 9 \to 16 = 2x \to x = 8$ . Check: $3(8) + 7 = 31$ and $5(8) - 9 = 31$ . On a graph, $x = 8$ is where the two sides meet.

y = 3x + 7

and

y = 5x - 9

meet at

(8, 31)

: both sides of

3x + 7 = 5x - 9

equal 31 when

x = 8

PredictRead the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.

When do they cost the same?

Try each option in both sides. Or solve it: what could you take from both sides first?

Which value of

x

makes

3x + 7 = 5x - 9

true?

x = 2

x = -8

x = 1

x = 8

TryTry it on the board and compare the result with your prediction.

Where both sides are equal

Read $P$ 's coordinates. Which number is the solution, and what does the other one tell you?

Try it yourselfYour edits are saved in this browser.

The two lines are $y = 3x + 7$ and $y = 5x - 9$ . Drag $P$ to where they meet. Was your prediction right?

Use the board. This ticks when you get there.

InvestigateWork out why it happens, one step at a time.

Same to both sides

Here is how to solve with $x$ on both sides.

Worked exampleBefore each step, say what you think comes next.

$3x + 7 = 5x - 9$ $x$ is on both sides.
$7 = 2x - 9$ Take $3x$ from both sides (the smaller $x$ term, so $2x$ stays positive).
$16 = 2x$ Add 9 to both sides: this undoes $- 9$ .
$x = 8$ Divide both sides by 2.
$3(8) + 7 = 31,\; 5(8) - 9 = 31$ Check in the first equation: both sides are 31.

Now solve it yourself, one step per line. The checker marks the first line that does not follow.

Read Ben's working one line at a time. How do you undo $- 7$ ?

Solve $3x + 7 = 5x - 9$ . Write one step per line.

Start from $3x + 7 = 5x - 9$

Ben solves $2x + 5 = 4x - 7$ . Click the line where his working goes wrong.

Check $x = 8$ in $3x + 7 = 5x - 9$ : what is the value of each side?

ModifyChange the problem on purpose and see what happens.

One change at a time

Each task changes one thing. Check each answer by substituting.

A bracket: solve $2(x + 3) = x + 10$ . Expand first. One step per line.

Start from $2(x + 3) = x + 10$

A negative answer: solve $4x - 3 = 7x + 12$ .

A fraction: solve $\dfrac{x}{3} + 2 = 6$ .

Brackets on both sides: solve $3(x - 2) = 2(x + 1)$ . One step per line.

Start from $3(x - 2) = 2(x + 1)$

Edge case: solve

2x + 3 = 2x + 5

. What happens?x = 0x = 1No value works: taking 2x away leaves 3 = 5, which is never true

PractiseA few on your own, to check you have it.

Solve and check

Solve $5x - 4 = 2x + 11$ . One step per line.

Start from $5x - 4 = 2x + 11$

Solve $6 - x = 2x - 9$ .

Angles on a straight line are $3x$ and $x + 20$ degrees. Find $x$ .

Review: expand $4(2x - 3)$ .

MakeMake your own problem using what you have learned. Expect a few tries; that is normal.

Build an equation

Write your equation, then solve it on paper to check.

Try it yourselfYour edits are saved in this browser.

Need a hint?

At x = 3, the first line is at y = 7. Your line must pass through (3, 7).

Another hint

Try m = 4: then c must make 4 × 3 + c = 7.

Set $m$ and $c$ so that $2x + 1 = mx + c$ has the solution $x = 3$ , with $m$ more than 2.

Use the board. This ticks when you get there.

For teachers

Students solve linear equations with x on both sides, brackets, negative answers and fractions by doing the same to both sides, writing one step per line, and verify every answer by substitution; a graph shows the solution as where the two sides meet (AC9M8A02, node 2). About 50 minutes.

Open the teacher guide: answers, solutions, slides and a worksheet.

Students will be able to

I can solve an equation with x on both sides, including negatives.
I can solve an equation with brackets by expanding first.
I can check my solution by substituting it into the first equation.

Before class

Do the working check yourself first. Students need The balance method (Year 7), Expanding brackets, and adding and multiplying negatives.

This is a hinge node: inequalities (next) use the same moves, and sign slips carry over. The common mistake is undoing a term with its own sign (taking 7 away to undo − 7); it is planted in Ben's working and again in Practise.

The working check marks each line against the one before, so students see the first step that went wrong; the last line must be x = …

Running the lesson

Predict: take a show of hands, then ask how students chose: trying values is fine, and leads to checking.
Run: ask what P's y-coordinate means (the value of both sides).
Investigate: model the worked example, saying why the smaller x term moves (so x stays positive).
Modify: for the no-solution task, ask what the graph would look like.
Make: collect students' equations and solve one together.

When students get stuck

A student takes 7 away to undo − 7 (Ben's mistake). Ask: What undoes taking 7 away? Fix: Use the inverse: add 7 to both sides.
A student leaves x on both sides after the first step. Ask: Which x term will you remove from both sides? Fix: Take the smaller x term from both sides first.
A student expands only the first term of a bracket. Ask: What does the number outside multiply? Fix: Every term inside: 2(x + 3) = 2x + 6.

Stretch and support

Stretch: Solve (x + 4)/2 = (x − 1)/3. (Multiply both sides by 6: 3x + 12 = 2x − 2, x = −14.)

Support: Use the Run graph to check answers: drag P to the meeting point.

Where it leads

Next: Inequalities (node 3), which solve the same way except when multiplying or dividing by a negative.

Where it leads: simultaneous equations (Year 9 and 10), and rearranging formulas with the subject in two places.

What each step is for

when do they cost the same?: Committing to a solution of 3x + 7 = 5x − 9 before solving shows whether they can reason about both sides; the graph then shows the solution as where the two sides are equal.
where both sides are equal: Each side of the equation is a line; the solution is where they meet, because there both sides have the same value.
same to both sides: Move the smaller x term (so the x term stays positive), then the numbers, then divide; substitute to check. Ben's mistake takes 7 away to undo − 7, instead of adding it.
one change at a time: One change at a time: a bracket, a negative answer, a fraction, brackets on both sides, then the edge case of an equation with no solution.
solve and check: Equations with x on both sides, negatives and a geometry context, mixed; the inverse-sign slip is planted again; the review keeps expanding fresh.
build an equation: Choosing a second line that meets y = 2x + 1 at x = 3 builds an equation with that solution, reversing the lesson: from the answer to the equation.

Now you can

Solve an equation with x on both sides, including negatives.
Solve an equation with brackets by expanding first.
Check my solution by substituting it into the first equation.

Practise this skillNew questions, new numbers each timeGet 3 right in a row to mark it practised.Next lesson14. InequalitiesSolve an inequality like an equation, doing the same to both sides, but when you multiply or divide by a negative, reverse the inequality sign. Show the answer on a number line: open circle for < and >, closed for ≤ and ≥. Check with a value.

Source: content/maths/algebra/solving-equations/lesson.md