4.03 Two-step equations
We have already seen how to solve one-step equations. Many of these equations were solvable by inspection, so let's look at some equations which require more steps to solve.
A two-step equation is one that will require two steps to solve. They generally have a multiplication/division and a subtraction/addition. The following are all two-step equations.
$2x+5=11$2x+5=11 $\frac{1}{3}h-3=6$13h−3=6 $\frac{9-j}{2}=5$9−j2=5
Let's look at two approaches to solving two-step equations. First, using algebra tiles and secondly, purely algebraically.
Using algebra tiles
Algebra tiles allow us to represent an equation more visually. It is important to ensure that you are keeping the two sides of the equation balanced, so what you do to one side, you must do to the other.
Exploration
Let's look at solving $2x-3=5$2x−3=5 using algebra tiles.
First, we need to set up the two sides of the equations, remember that red tiles represent negative values.
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$=$= |
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| $2x$2x | $-3$−3 | $=$= | $5$5 |
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Keeping the two sides balanced we want to add or remove tiles and work towards a single $x$x tile. Let's start by adding $3$3 positive unit tiles to both sides.
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| $2x$2x | $-3+3$−3+3 | $=$= | $5$5 |
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$+3$+3 |
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| $2x$2x | $=$= | $8$8 |
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| $2x$2x | $=$= | $8$8 |
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$=$= |
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| $x$x | $=$= | $4$4 |
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Using formal algebraic techniques
If we don't have algebra tiles available or if we have an equation involving fractions, then solving purely algebraically is also an option. Remember from when we solved one step equations:
- When asked to solve, we want to get the variable by itself on one side of the equals sign
- To keep everything balanced, we must do the same operations to both sides by applying the properties of equality
- We should use opposite operations to solve
- Addition and subtraction are opposite operations
- Multiplication and division are opposite operations
Worked examples
Question 1
Solve for $x$x in $-2x+4=8$−2x+4=8, showing all of your work algebraically.
Think: There are two operations happening to the $x$x, we are multiplying by $-2$−2 and adding $4$4. Which should we reverse first? We should go in the opposite order they were done in, so we will first subtract $4$4.
Do:
| $-2x+4$−2x+4 | $=$= | $8$8 |
The opposite of addition is subtraction. |
| $-2x+4-4$−2x+4−4 | $=$= | $8-4$8−4 |
Start by subtracting $4$4 from both sides. |
| $-2x$−2x | $=$= | $4$4 |
Simplify both sides of the equation |
| $\frac{-2x}{-2}$−2x−2 | $=$= | $\frac{4}{-2}$4−2 |
The opposite of multiplying by $-2$−2 is dividing by $-2$−2, so do this to both sides. |
| $x$x | $=$= | $-2$−2 |
Simplify both sides to find $x$x |
Reflect: It is always a good idea to check your answer by substituting it back in to the original equation. Does $-2\left(-2\right)+4=8$−2(−2)+4=8?
Question 2
Solve for $x$x in $\frac{x}{3}-2=15$x3−2=15, showing all of your work algebraically.
Think: There are two operations happening to the $x$x, we are dividing by $3$3 and subtracting $2$2. Which should we reverse first? We should go in the opposite order they were done in, so we will first add $2$2 first.
Do:
| $\frac{x}{3}-2$x3−2 | $=$= | $15$15 |
The given equation |
| $\frac{x}{3}-2+2$x3−2+2 | $=$= | $15+2$15+2 |
Start by adding $2$2 to both sides |
| $\frac{x}{3}$x3 | $=$= | $17$17 |
Simplify both sides |
| $3\frac{x}{3}$3x3 | $=$= | $3\times17$3×17 |
Multiply both sides by $3$3 |
| $x$x | $=$= | $51$51 |
Simplify both sides to find $x$x |
| $\frac{x}{3}-2$x3−2 | $=$= | $15$15 |
The given equation |
| $\frac{x}{3}-2+2$x3−2+2 | $=$= | $15+2$15+2 |
Start by adding $2$2 to both sides |
| $\frac{x}{3}$x3 | $=$= | $17$17 |
Simplify both sides |
| $3\frac{x}{3}$3x3 | $=$= | $3\times17$3×17 |
Multiply both sides by $3$3 |
| $x$x | $=$= | $51$51 |
Simplify both sides to find $x$x |
Reflect: Is $x=51$x=51 a reasonable answer? Does $\frac{51}{3}-2=15$513−2=15?
Practice questions
Question 3
Solve the following equation: $8x-9=39$8x−9=39
Question 4
Solve the following equation: $-10+3k=5$−10+3k=5
Question 5
Solve the following equation: $\frac{x}{2}+8=10$x2+8=10