Solving Two-Step Equations: math lesson, Grade 7

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Math lessons Grade 7 : Solving Two-Step Equations — Zyro the alien explorer of Planète Maths

A one-step equation needs a single move to find the unknown. Many real situations hide two operations at once: a fee plus a price per item, a starting amount plus a weekly saving, a temperature conversion. In this chapter you will learn a reliable routine for undoing both operations, writing equations from word problems, and proving that your answer is right.

Sommaire

1. What is a two-step equation?

An equation says that two expressions have the same value. In a two-step equation the variable has been changed by two operations, so you need two inverse moves to free it. You will meet two main shapes:

$px + q = r$, for example $3x + 4 = 19$: the variable is multiplied, and then a number is added or subtracted;
$p(x + q) = r$, for example $4(x + 3) = 44$: a number is added to the variable first, and then the whole sum is multiplied.

Definition: solutionA solution of an equation is a value of the variable that makes the equation true. To solve an equation means to find all its solutions. Each two-step equation in this chapter has exactly one.

2. Inverse operations and the balance idea

Think of an equation as a balanced scale. Whatever you do to one side, you must do to the other, or the balance tips. Each operation has an inverse operation that undoes it.

Operation done to the variable	Inverse operation that undoes it
Add a number, $x + 7$	Subtract that number
Subtract a number, $x - 7$	Add that number
Multiply by a number, $7x$	Divide by that number
Divide by a number, $\dfrac{x}{7}$	Multiply by that number

Look at the bar model for $3x + 4 = 19$. Three equal boxes of size $x$ and one box of 4 together make 19.

Take away the box of 4 from the total: three boxes of $x$ make $19 - 4 = 15$. Share 15 into 3 equal boxes: each $x$ is 5.

3. Solving $px + q = r$

Property: equal operationsAdding, subtracting, multiplying or dividing both sides of an equation by the same number (never zero for multiplying or dividing) gives an equation with the same solution. This is why inverse operations are allowed.

Method: solve px + q = r

Undo the addition or subtraction first: subtract $q$ from both sides (or add it if $q$ is negative).
You now have $px = r - q$. Divide both sides by $p$.
Write the answer as $x = \dots$.
Check by replacing $x$ in the original equation.

Example 1Solve $3x + 4 = 19$.
Subtract 4 from both sides: $3x = 15$.
Divide both sides by 3: $x = 5$.
Check: $3(5) + 4 = 15 + 4 = 19$. It works.

Example 2Solve $5x - 9 = 26$.
Add 9 to both sides: $5x = 35$.
Divide by 5: $x = 7$.
Check: $5(7) - 9 = 35 - 9 = 26$. It works.

Common mistakeDo not divide only one term. If you divide the equation $3x + 6 = 21$ by 3 right away, you must divide every term: $x + 2 = 7$. Subtracting first is usually easier and safer.

4. Working backward with a flow chart

A flow chart shows what the equation does to $x$, step by step. To solve, run the chart backward: start from the result, and apply the inverse operations in reverse order.

Forward, $x$ is multiplied by 3 and then increased by 4. Backward, 19 is decreased by 4 and then divided by 3. The two routes always meet in the same answer.

Zyro’s tipOn my home planet we get dressed in order and undress in the opposite order. Equations work the same way: the last operation done to $x$ is the first one you undo!

5. Solving $p(x + q) = r$

When a sum or difference sits inside parentheses and is multiplied, you have two valid routes.

Method: solve p(x + q) = r

Route A (divide first): divide both sides by $p$, then subtract $q$.
Route B (distribute first): expand to $px + pq = r$, then solve like $px + q = r$.
Check in the original equation.

Example 3Solve $6(x - 2) = 42$.
Route A: divide by 6: $x - 2 = 7$. Add 2: $x = 9$.
Route B: $6x - 12 = 42$, so $6x = 54$ and $x = 9$.
Check: $6(9 - 2) = 6 \times 7 = 42$. Both routes agree.

Route A is faster when $p$ divides $r$ evenly. If it does not, as in $4(x + 1) = 10$, the answer is a fraction or decimal: $x + 1 = 2.5$, so $x = 1.5$.

6. Rational coefficients

The numbers $p$, $q$ and $r$ may be fractions, decimals or negative numbers. The routine does not change.

Divided by a number: to undo $\dfrac{x}{4}$, multiply by 4.
Fraction coefficient: to undo $\dfrac{2}{5}x$, multiply by the reciprocal $\dfrac{5}{2}$.
Decimal coefficient: divide by the decimal, for example $2.5x = 10$ gives $x = 4$.
Negative coefficient: divide by the negative number, and watch the sign: $-3x = -18$ gives $x = 6$.

Example 4Solve $\dfrac{x}{4} - 3 = 2$.
Add 3: $\dfrac{x}{4} = 5$. Multiply by 4: $x = 20$.
Check: $\dfrac{20}{4} - 3 = 5 - 3 = 2$.

Example 5Solve $\dfrac{2}{5}x + 1 = 7$.
Subtract 1: $\dfrac{2}{5}x = 6$. Multiply by $\dfrac{5}{2}$: $x = 6 \times \dfrac{5}{2} = 15$.
Check: $\dfrac{2}{5}(15) + 1 = 6 + 1 = 7$.

7. Writing equations from word problems

Method: from words to an equation

Read the problem and say what the unknown is. Write “Let $m$ = number of months”.
Find the fixed amount (paid once or at the start) and the amount that repeats.
Write the expression: repeating amount times the variable, plus or minus the fixed amount.
Set it equal to the total given, solve, and answer with a full sentence and units.

Example 6A video app charges $5 to join and $8 per month. Kai has paid $61 in total. How many months has he subscribed?
Let $m$ be the number of months: $8m + 5 = 61$.
Subtract 5: $8m = 56$. Divide by 8: $m = 7$.
Kai has subscribed for 7 months. Check: $8(7) + 5 = 61$.

Watch the key phrases: “more than” and “plus” point to addition, “per” points to multiplication, and “twice” means times 2. “7 more than twice a number is 25” becomes $2n + 7 = 25$, not $2(n + 7) = 25$.

8. Checking and comparing methods

Always check a solution by substituting it into the original equation and confirming that both sides are equal. On a number line the solution is a single point, here $x = 5$:

You can also see a solution on a graph. The line $y = 2x + 3$ reaches the height 11 exactly when $x = 4$, so $2x + 3 = 11$ has the solution $x = 4$:

An arithmetic solution works backward with numbers only. An algebraic solution writes an equation and uses inverse operations. For “I think of a number, double it and add 5. I get 21”, the arithmetic way is $21 - 5 = 16$, then $16 \div 2 = 8$. The algebraic way is $2x + 5 = 21$, $2x = 16$, $x = 8$. The steps are identical; algebra simply gives them a clear written form that also works for harder problems.

Key takeaways

A two-step equation has two operations on the variable, so it needs two inverse operations.
Undo in reverse order: first add or subtract, then multiply or divide.
For $px + q = r$: $x = \dfrac{r - q}{p}$. For $p(x + q) = r$: divide by $p$ first, or distribute first.
Do the same thing to both sides to keep the equation balanced.
Fractions, decimals and negative numbers follow the same routine; use the reciprocal for a fraction.
For word problems, define the variable, build the equation, solve, then answer in a sentence with units.
Always check by substituting in the original equation.