Math 7 · 7.EE.B.4A · 8.EE.C.7

Two-step equations

A number was multiplied, then something was added, and only the result is known. How do you get back to the number?

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▴ 1 above Inverse operations and the properties of equality One-step equations

Two-step equations

Equations from proportions and percents Equations of the form p(x + q) = r, from words ▾ Show all 11 (5 more)

When you reach for it

A one-step equation such as x + 5 = 26 or 3x = 21 has one thing done to the unknown, and one move undoes it. Many equations have two things done to it:

3x + 5 = 26

Here x was multiplied by 3, and then 5 was added. An equation of this shape is a two-step equation:

px + q = r

where p, q and r are numbers, p is not 0, and the unknown appears exactly once. The number p that multiplies x is its coefficient, and q is the constant term. The numbers may be negative, fractions or decimals. An equation such as \displaystyle \frac{m}{5} + 2 = 9 has the same shape, since dividing by 5 is multiplying by \displaystyle \frac{1}{5}.

You reach for this skill whenever an unknown has been scaled and then shifted: so much per visit plus a fixed charge, a steady rate of climb or descent from a starting height, points per level less a penalty. The question is always the same one, run backwards: what number went in, if this came out?

This lesson takes three days. Part 1 solves px + q = r with positive numbers, with a bar picture to show why each step works. Part 2 lets the coefficient and the constant be negative, fractions or decimals. Part 3 writes the equation from a story, solves it, and checks the answer in the story.

The procedure

Part 1: undoing in reverse order

Read 3x + 5 = 26 as a record of what happened to a number. By the order of operations, the multiplication comes first and the addition second: start with x, multiply by 3, add 5, and the result is 26.

To get back to x, undo those steps in reverse order: the last step done is the first one undone. It is the order you take off a coat and then the jumper under it.

Figure 1The equation records what was done to x. Undoing runs the same chain backwards, last step first.

Undo the addition: subtract 5 from both sides. The subtraction property of equality says that taking the same number from both sides of a true equation leaves it true. On the left, +5 and -5 cancel:

3x + 5 - 5 = 26 - 5 \qquad\text{so}\qquad 3x = 21

Undo the multiplication: divide both sides by 3. The division property of equality allows dividing both sides by the same number, provided it is not 0:

\frac{3x}{3} = \frac{21}{3} \qquad\text{so}\qquad x = 7

Check by putting 7 back into the original equation, which is what makes it a solution: 3(7) + 5 = 21 + 5 = 26. It works.

A bar picture shows the same two steps without symbols. The whole bar is 26 long. It is made of three equal pieces, each x long, and one piece 5 long.

Figure 2The bar model of 3x + 5 = 26. Take away the 5 and the three equal pieces share 21.

Taking away the 5 piece leaves 26 - 5 = 21 for the three equal pieces, and three equal pieces sharing 21 are each 7. That is subtracting first and dividing second, for a reason you can see.

Why there is always exactly one answer. Each step can be undone again (add 5 back, multiply by 3 again), so the new equation has exactly the same solutions as the old one. The last equation, x = 7, has one solution, so the first one did too. In general, subtracting q and dividing by p turns px + q = r into

x = \frac{r - q}{p}

which is one number, provided p \neq 0.

Reverse order is the tidiest route, not the only one. Dividing by 3 first is allowed too, but then every term on both sides is divided, by the distributive property, and fractions appear: \displaystyle x + \frac{5}{3} = \frac{26}{3}. Example 3 does it both ways.

Part 2: negative numbers, fractions and decimals

Nothing in the procedure needs the numbers to be positive. What changes is reading the equation.

A minus sign belongs to the term after it. Since subtracting is adding the opposite, the equation 4 - 3k = -11 is 4 + (-3k) = -11. Its coefficient is -3, not 3, and its constant is 4: the unknown was multiplied by -3, then 4 was added. So subtract 4 from both sides, then divide both sides by -3 (Example 4 finishes it). Dividing by a negative number is allowed, since the division property of equality needs only a non-zero divisor, and the signs follow the rules for multiplying and dividing signed numbers. An unknown written -x has coefficient -1.

A fraction coefficient is undone by its reciprocal. If the unknown was multiplied by \displaystyle \frac{2}{3}, dividing by \displaystyle \frac{2}{3} undoes it, and dividing by a fraction is multiplying by its reciprocal: multiply both sides by \displaystyle \frac{3}{2}. This works because \displaystyle \frac{3}{2} \cdot \frac{2}{3} = 1, which leaves 1x = x.

Or clear the fractions first. Multiplying both sides by a common denominator of every fraction in the equation gives an equivalent equation with whole-number coefficients. The multiplication reaches every term, by the distributive property. Example 5 shows both routes.

Decimals are fractions with denominators 10, 100 and so on. Either divide by the decimal directly, or multiply every term by 10 or 100 to clear the decimal points first.

Part 3: writing the equation from words

A story does not hand you the equation. You build it, and the building is most of the work.

Name the unknown, with its unit. “Let v be the number of visits” is a full sentence and worth writing, because the equation, the answer and the check all have to agree about what v counts.

Find the quantity the story tells you two ways. In a two-step story it is usually an end amount: once as an expression built from the start, the rate and the unknown, and once as a number. Set the two equal. A steady change of p per step for x steps, from a start of q, ends at px + q. If the amount goes down, the change per step is negative. The work of turning phrases into expressions is translating words to expressions, and a bar model is a good way to see what equals what.

Solve, then check in the story, not only in the equation. Putting the answer back into your own equation catches arithmetic slips, but not an equation built wrongly. Reading the answer back into the story, in its own words and units, catches both. Then ask whether the number can happen: a count of visits or levels must be a whole number, and a time must not be negative.

Worked examples

Examples 1 to 3 go with Part 1. Examples 4 to 6 go with Part 2. Examples 7 to 9 go with Part 3.

Example 1

Undoing two steps

Solve (a) 5a + 12 = 47 \qquad (b) 33 = 4n + 9

Solution

(a) The unknown was multiplied by 5, then 12 was added. Undo the addition first: subtract 12 from both sides. Then undo the multiplication: divide both sides by 5.

\begin{aligned} 5a + 12 &= 47 \\ 5a &= 35 \\ a &= 7 \end{aligned}

Check: 5(7) + 12 = 35 + 12 = 47.

(b) The unknown is on the right, which changes nothing: an equation says its two sides are equal, and 4n + 9 = 33 says exactly the same thing (the symmetric property of equality). Subtract 9 from both sides, then divide both sides by 4:

\begin{aligned} 33 - 9 &= 4n \\ 24 &= 4n \\ 6 &= n \end{aligned}

So n = 6. Check: 4(6) + 9 = 24 + 9 = 33.

Example 2

Division as the first step

Solve (a) \displaystyle \frac{m}{5} + 2 = 9 \qquad (b) \displaystyle 2 + \frac{c}{6} = 10

Solution

(a) Here m was divided by 5, then 2 was added. Undo in reverse order: subtract 2, then undo the division by multiplying both sides by 5.

\begin{aligned} \frac{m}{5} &= 7 \\ m &= 35 \end{aligned}

Check: \displaystyle \frac{35}{5} + 2 = 7 + 2 = 9.

(b) The constant is written first, but addition is commutative, so \displaystyle 2 + \frac{c}{6} is the same as \displaystyle \frac{c}{6} + 2. The order of the steps done to c is still divide by 6, then add 2. Subtract 2 from both sides, then multiply both sides by 6:

\frac{c}{6} = 8 \qquad\text{so}\qquad c = 48

Check: \displaystyle 2 + \frac{48}{6} = 2 + 8 = 10.

Example 3

A fraction answer, and the other order

Solve 6y - 8 = 19, first in reverse order and then by dividing first.

Solution

The unknown was multiplied by 6, then 8 was subtracted. Undo the subtraction by adding 8 to both sides, then divide both sides by 6:

\begin{aligned} 6y &= 27 \\ y &= \frac{27}{6} = \frac{9}{2} \end{aligned}

The answer is not a whole number, and need not be. Check: \displaystyle 6 \cdot \frac{9}{2} = 27, and 27 - 8 = 19.

Now divide first. Dividing both sides by 6 divides every term on the left, because division distributes over subtraction:

\begin{aligned} \frac{6y}{6} - \frac{8}{6} &= \frac{19}{6} \\ y - \frac{4}{3} &= \frac{19}{6} \\ y &= \frac{19}{6} + \frac{8}{6} = \frac{27}{6} = \frac{9}{2} \end{aligned}

The same answer, as it must be, since every step was allowed. Reverse order simply keeps the fractions away until the last step.

Example 4

Negative coefficients and constants

Solve (a) -5x + 3 = 23 \qquad (b) 4 - 3k = -11 \qquad (c) -x - 8 = 2

Solution

(a) The coefficient is -5. Subtract 3 from both sides, then divide both sides by -5. A positive divided by a negative is negative:

-5x = 20 \qquad\text{so}\qquad x = \frac{20}{-5} = -4

Check: -5(-4) + 3 = 20 + 3 = 23.

(b) Read 4 - 3k as 4 + (-3k): the coefficient is -3 and the constant is 4. Subtract 4 from both sides, then divide both sides by -3:

-3k = -15 \qquad\text{so}\qquad k = \frac{-15}{-3} = 5

The solution is positive even though the equation is full of minus signs. Check: 4 - 3(5) = 4 - 15 = -11.

-x = 10 \qquad\text{so}\qquad x = -10

Check: -(-10) - 8 = 10 - 8 = 2.

Example 5

Fraction coefficients

Solve (a) \displaystyle \frac{2}{3}x - 5 = 7 \qquad (b) \displaystyle \frac{3}{4}n + \frac{1}{2} = -1

Solution

(a) Add 5 to both sides, then undo the multiplication by \displaystyle \frac{2}{3} by multiplying both sides by its reciprocal, \displaystyle \frac{3}{2}:

\begin{aligned} \frac{2}{3}x &= 12 \\ x &= \frac{3}{2} \cdot 12 = 18 \end{aligned}

Check: \displaystyle \frac{2}{3} \cdot 18 = 12, and 12 - 5 = 7.

(b) The denominators are 4 and 2, and 4 is a multiple of both, so multiply every term of both sides by 4. That gives \displaystyle 4 \cdot \frac{3}{4}n = 3n and \displaystyle 4 \cdot \frac{1}{2} = 2:

\begin{aligned} 3n + 2 &= -4 \\ 3n &= -6 \\ n &= -2 \end{aligned}

Working in reverse order without clearing gives the same: subtracting \displaystyle \frac{1}{2} gives \displaystyle \frac{3}{4}n = -\frac{3}{2}, and multiplying by \displaystyle \frac{4}{3} gives n = -2. Check: \displaystyle \frac{3}{4}(-2) + \frac{1}{2} = -\frac{3}{2} + \frac{1}{2} = -1.

Example 6

Decimals, and a negative fraction

Solve (a) 0.4w - 1.5 = 2.3 \qquad (b) \displaystyle 5 - \frac{2}{5}p = 9

Solution

(a) Every decimal here has one decimal place, so multiplying every term by 10 clears them all:

\begin{aligned} 4w - 15 &= 23 \\ 4w &= 38 \\ w &= 9.5 \end{aligned}

Dividing directly gives the same: adding 1.5 gives 0.4w = 3.8, and 3.8 \div 0.4 = 9.5. Check: 0.4 \times 9.5 = 3.8, and 3.8 - 1.5 = 2.3.

(b) The coefficient is \displaystyle -\frac{2}{5} and the constant is 5. Subtract 5 from both sides, then multiply both sides by the reciprocal of \displaystyle -\frac{2}{5}, which is \displaystyle -\frac{5}{2}:

\begin{aligned} -\frac{2}{5}p &= 4 \\ p &= -\frac{5}{2} \cdot 4 = -10 \end{aligned}

The reciprocal of a negative number is negative, since their product must be +1. Check: \displaystyle 5 - \frac{2}{5}(-10) = 5 + 4 = 9.

Example 7

A gift card

Ulyana has a café gift card worth 60 dollars. On each visit she buys the same drink for 4.50 dollars, paid from the card. After some visits the card shows 28.50 dollars left. How many visits has she made?

Solution

Let v be the number of visits. The amount left on the card is told two ways. Built from the story, it is the starting 60 less 4.50 for each of the v visits; and it is 28.50. So

60 - 4.5v = 28.5

Subtract 60 from both sides, then divide both sides by -4.5:

\begin{aligned} -4.5v &= -31.5 \\ v &= \frac{-31.5}{-4.5} = 7 \end{aligned}

Check in the story. Seven drinks at 4.50 dollars cost 31.50 dollars, and 60 - 31.50 = 28.50 dollars is left. Seven is a whole number of visits, so the answer makes sense: 7 visits.

The arithmetic route is the same reasoning: 60 - 28.50 = 31.50 dollars was spent, and 31.50 \div 4.50 = 7. Undoing the subtraction first, then the multiplication. The equation keeps that reasoning in order when the numbers are awkward or negative.

Example 8

Down into a canyon

Wilhelm measures heights on a canyon trail from the rim, so points below the rim have negative heights. He starts at a viewpoint 40 m below the rim and walks down at a steady 12 m each minute. After how many minutes does he reach a ledge 220 m below the rim?

Solution

Let t be the number of minutes. His height starts at -40 and changes by -12 each minute, so after t minutes it is -40 + (-12)t, which is -40 - 12t. The ledge is at -220:

-40 - 12t = -220

Add 40 to both sides, then divide both sides by -12:

\begin{aligned} -12t &= -180 \\ t &= 15 \end{aligned}

Figure 3Heights in Example 8, measured from the rim. Going down makes the height more negative.

Check in the story. In 15 minutes at 12 m a minute he goes down 180 m, and 40 m below the rim plus another 180 m is 220 m below it. He reaches the ledge after 15 minutes.

The negative heights did the bookkeeping. Had the descent been written +12t, the equation would have given t = -15, and a negative time is the story’s signal that a sign is wrong.

Example 9

Points in a phone game

In a phone game, each level cleared earns 150 points and each hint used costs 40 points. Zuleika starts at 0 points and uses 6 hints.

(a) She finishes with 1{,}110 points. How many levels did she clear?

(b) Could she have finished with exactly 1{,}000 points, still using 6 hints?

Solution

(a) Let L be the number of levels. The levels give 150L points and the hints take away 40 \times 6 = 240, so her total is 150L - 240:

\begin{aligned} 150L - 240 &= 1110 \\ 150L &= 1350 \\ L &= 9 \end{aligned}

Check in the story. Nine levels earn 1{,}350 points; six hints cost 240; 1{,}350 - 240 = 1{,}110. She cleared 9 levels.

(b) The same equation with 1{,}000 on the right:

\begin{aligned} 150L - 240 &= 1000 \\ 150L &= 1240 \\ L &= \frac{1240}{150} = \frac{124}{15} \end{aligned}

The equation has a solution, \displaystyle 8\frac{4}{15}, but the story does not: a number of levels is a whole number. Eight levels give 1{,}200 - 240 = 960 points and nine give 1{,}110, so with 6 hints no number of levels lands on 1{,}000. No. The algebra answered the equation; the story decided whether that answer could happen.

Boundaries

The coefficient must not be 0. The last step divides by p. If p = 0 there is no x left to find: 0x + 3 = 5 is false for every number, and 0x + 3 = 3 is true for every number. Telling those cases apart is its own lesson.

The unknown must appear once. Two-step equations have one x term. If x appears in two terms on the same side, they are combined first, as in multi-step equations. If it appears on both sides, it is gathered onto one side first. If the unknown sits inside brackets, as in p(x + q) = r, the steps are different and belong to equations of the form p(x + q) = r.

A move must reach every term of both sides. Adding or subtracting changes both sides by the same amount. Multiplying or dividing both sides multiplies or divides each term in them, by the distributive property; clearing fractions from only some terms changes the equation.

Dividing by a negative is fine in an equation, not in an inequality as it stands. Equality survives division by any non-zero number. An inequality reverses when both sides are multiplied or divided by a negative, which is the extra care in two-step inequalities.

A story can reject an answer the equation accepts. The equation knows nothing about visits, levels or minutes. A count must be a whole number, a time or a load cannot be negative, and an answer that breaks those rules means either no such situation exists, as in Example 9(b), or the equation was built with a wrong sign.

Where this goes next

Every longer linear equation is solved by turning it into a two-step equation. Multi-step equations tidy each side by combining like terms first, equations with the unknown on both sides gather the x terms onto one side, and equations of the form p(x + q) = r either divide first or distribute first. Equations from proportions reach the same shape after cross multiplying.

The same undoing, done with letters in place of numbers, is solving for a variable: px + q = r becomes \displaystyle x = \frac{r - q}{p}. With the equals sign replaced by < or >, the steps carry over to two-step inequalities, with one rule added for negatives. And the story shape of Part 3, a start plus a steady rate times an amount, is a linear function; modeling with linear relationships studies the whole line, where this lesson asked for one point on it.

Exercises

Part 1: undoing two steps

Solve each equation and check your answer.

2x + 9 = 23
30 = 4p + 6
\displaystyle \frac{a}{3} + 8 = 14
8k - 5 = 25
A bar 45 long is cut into four equal pieces of length x and one piece of length 13. Write an equation for the bar and find x.

Part 2: signed numbers

-3x + 7 = 22
6n - 11 = -47
12 - 5y = -8
-x + 9 = -3
-14 = -2t - 20
\displaystyle \frac{a}{-4} - 3 = 2

Part 2: fractions and decimals

\displaystyle \frac{3}{5}x + 4 = 10
\displaystyle \frac{1}{2}y - \frac{3}{4} = \frac{5}{4}
\displaystyle -\frac{2}{3}k + 1 = 9
0.3m + 2.1 = 4.5
1.25 - 0.5d = -2.25
\displaystyle \frac{5}{6}z - \frac{2}{3} = -3

Part 3: from words

Write an equation, solve it, and check the answer in the story.

Yannick’s drone hovers 95 m above a park, then descends at a steady 2.5 m each second. After how many seconds is it 30 m above the park?
A delivery van leaves the depot carrying 860 kg of parcels and unloads 65 kg at each stop. After how many stops is its load 340 kg?
After a penalty, Valentin’s score in a phone game is -50 points. Each coin he collects is worth 15 points. How many coins must he collect to reach 400 points?
Heights in a mine are measured from ground level. A maintenance cage is at -180 m and is hauled up at a steady 15 m each minute. After how many minutes is it at -45 m?
A van carries 900 kg and unloads 70 kg at each stop. Could its load ever be exactly 250 kg? Explain.
Write a short story, in one of the settings above, that the equation -4x + 30 = 6 describes. Then solve the equation and say what the answer means in your story.

Reasoning

Solve 4x + 10 = 30 by dividing both sides by 4 first. Explain why the 10 is divided too.
Use the bar model or the chain of operations to explain why, for 2x + 9 = 23, the 9 is dealt with before the 2.
Explain why multiplying both sides of \displaystyle \frac{2}{3}x = 12 by \displaystyle \frac{3}{2} gives the same result as dividing both sides by \displaystyle \frac{2}{3}.
Show that px + q = r with p \neq 0 has exactly one solution, \displaystyle x = \frac{r - q}{p}. What happens when p = 0?

Answers

x = 7. Subtract 9: 2x = 14. Divide by 2.
p = 6. Subtract 6: 24 = 4p. Divide by 4.
a = 18. Subtract 8: \displaystyle \frac{a}{3} = 6. Multiply by 3.
\displaystyle k = \frac{15}{4}, which is 3.75. Add 5: 8k = 30. Divide by 8.
4x + 13 = 45, so x = 8. Take away the 13 piece: 4x = 32, and four equal pieces share 32.
x = -5. Subtract 7: -3x = 15. Divide by -3.
n = -6. Add 11: 6n = -36. Divide by 6.
y = 4. Subtract 12: -5y = -20. Divide by -5.
x = 12. Subtract 9: -x = -12. Divide by -1.
t = -3. Add 20: 6 = -2t. Divide by -2.
a = -20. Add 3: \displaystyle \frac{a}{-4} = 5. Multiply by -4.
x = 10. Subtract 4: \displaystyle \frac{3}{5}x = 6. Multiply by \displaystyle \frac{5}{3}.
y = 4. Multiply every term by 4: 2y - 3 = 5, so 2y = 8.
k = -12. Subtract 1: \displaystyle -\frac{2}{3}k = 8. Multiply by \displaystyle -\frac{3}{2}.
m = 8. Subtract 2.1: 0.3m = 2.4. Divide by 0.3.
d = 7. Subtract 1.25: -0.5d = -3.5. Divide by -0.5.
\displaystyle z = -\frac{14}{5}. Multiply every term by 6: 5z - 4 = -18, so 5z = -14. Check: \displaystyle \frac{5}{6} \cdot \left(-\frac{14}{5}\right) = -\frac{7}{3}, and \displaystyle -\frac{7}{3} - \frac{2}{3} = -3.
26 seconds. 95 - 2.5t = 30, so -2.5t = -65. Check: 2.5 \times 26 = 65 m down, and 95 - 65 = 30.
8 stops. 860 - 65s = 340, so -65s = -520. Check: 8 stops unload 520 kg, and 860 - 520 = 340.
30 coins. -50 + 15c = 400, so 15c = 450. Check: 30 coins are worth 450, and -50 + 450 = 400.
9 minutes. -180 + 15t = -45, so 15t = 135. Check: 9 minutes of hauling is 135 m, and -180 + 135 = -45.
No. 900 - 70s = 250 gives 70s = 650 and \displaystyle s = \frac{65}{7}, about 9.29, which is not a whole number of stops. After 9 stops the load is 270 kg and after 10 it is 200 kg.
x = 6. Subtract 30: -4x = -24; divide by -4. One story: a drone hovers 30 m up and descends 4 m each second; after x seconds it is 6 m up, which takes 6 seconds. Any story with a start of 30, a change of -4 per step and an end of 6 fits.
x = 5. Dividing both sides by 4 gives \displaystyle x + \frac{10}{4} = \frac{30}{4}, that is, x + 2.5 = 7.5, so x = 5. The whole left side, 4x + 10, is divided by 4, and by the distributive property that divides each of its terms. Dividing only the 4x would change the left side by a different amount from the right, and the equation would no longer say the same thing.
The left side was built from x by doubling and then adding 9, so the 9 is the outermost layer. In the bar, the 9 piece is separate from the two equal pieces, so it can be taken away at once, leaving 14 to be shared by two pieces of 7. Halving first would mean halving the 9 piece as well. Undoing the last step first gets back to 2x = 14, the state before 9 was added, and then dividing by 2 gets back to x.
Dividing by a non-zero number is multiplying by its reciprocal, and the reciprocal of \displaystyle \frac{2}{3} is \displaystyle \frac{3}{2}. Either way the left side becomes \displaystyle \frac{3}{2} \cdot \frac{2}{3}x = 1x = x and the right side becomes 18.
Subtract q from both sides: px = r - q. Divide both sides by p, allowed since p \neq 0: \displaystyle x = \frac{r - q}{p}. So any solution must be this number, and it does work: \displaystyle p \cdot \frac{r - q}{p} + q = r - q + q = r. When p = 0 the equation says q = r, with no x in it: if q = r every number is a solution, and if not, none is.