Math 7 · 6.EE.A.3 · 7.EE.A.1

Distribute, then combine

An expression has brackets with numbers in front, and terms of the same kind sit both inside and outside them. How do you bring it down to its shortest form without changing its value?

You need

The distributive property with variablesMath 7

Where you are 12 in Expressions

Algebra9

Expressions12

▴ 4 above The distributive property with variables Variables5

Distribute, then combine

Multiplying two binomials Rewriting an expression to show what it means ▾ Show all 12 (3 more)

When you reach for it

Gaspard is ordering food for a dinner. Each guest gets a meal costing 15 dollars and a drink costing 4 dollars. There were going to be g guests, but 3 have canceled, so he pays for g - 3 meals and g - 3 drinks, and there is a delivery charge of 30 dollars. The bill, in dollars, is

15(g - 3) + 4(g - 3) + 30

That is correct, and it follows the story piece by piece, but it hides the two things Gaspard most wants to know: how much each guest adds to the bill, and what is left over. Rewritten, the same bill is 19g - 27. Now it says that every guest adds 19 dollars, a meal and a drink, and the fixed part is -27: the 30 dollars of delivery minus the 3 \cdot 19 = 57 dollars the canceled guests no longer cost.

Getting from the first form to the second takes two moves you already have, done in a fixed order. First distribute: multiply the number in front of each bracket into every term inside it. Then combine like terms. You reach for this whenever an expression has brackets with numbers in front and you want its shortest equivalent form: before evaluating it, when comparing it with another expression, and above all when an equation with brackets in it has to be solved.

This lesson runs over two days. Part 1 says why the brackets are opened first, works expressions with one and two brackets, and shows how to check an answer by substituting a number. Part 2 brings in a negative number in front of a bracket, subtracting a whole bracket, fraction and decimal coefficients, and perimeters of shapes made from rectangles.

The procedure

Part 1: open the brackets, then collect

Why the brackets go first. In 3(x + 2) + 4x the x inside the bracket and the 4x outside are like terms, but they cannot be combined yet. The bracket says: work out x + 2, then multiply by 3. By the Order of operations, that multiplication happens before anything is added to it, so the 3 belongs to the whole bracket and nothing outside it can be gathered with the x while the bracket is closed. Distributing removes the bracket without changing the value, and then every term stands on its own:

\begin{aligned} 3(x + 2) + 4x &= 3x + 6 + 4x \\ &= 7x + 6 \end{aligned}

A picture shows what the first step does. Read each product as the area of a rectangle: 3(x + 2) is a rectangle 3 tall and x + 2 wide, and 2(x + 4) is one 2 tall and x + 4 wide. Distributing cuts each rectangle into an x-part and a number part, 3x + 6 and 2x + 8.

Figure 1Opened up, 3(x + 2) + 2(x + 4) is two cut rectangles. The x-parts line up into one column of 5x, and the plain numbers add to 6 + 8 = 14.

Stack the two and the x-parts line up in one column 3 + 2 = 5 tall and x wide, which is 5x. The number parts are two separate pieces of area, 6 and 8, which add to 14. So

3(x + 2) + 2(x + 4) = 5x + 14

Before the cut, the rectangles had different widths and nothing lined up. The like terms only appear once the brackets are open.

Open every bracket. Back to Gaspard’s bill. Each number in front multiplies every term inside its own bracket, signs included: 15(g - 3) = 15g - 45 and 4(g - 3) = 4g - 12. The 30 has no bracket, so it is copied as it is:

\begin{aligned} &15(g - 3) + 4(g - 3) + 30 \\ &= 15g - 45 + 4g - 12 + 30 \end{aligned}

Collect like terms. Now combine like terms: each sign travels with the term after it, and the commutative and associative properties of addition let the g terms be gathered together and the constants together:

\begin{aligned} &= (15g + 4g) + (-45 - 12 + 30) \\ &= 19g - 27 \end{aligned}

The answer is finished when no two terms are alike: one g term and one constant.

Check by substituting a number. The rewritten expression is meant to be equivalent to the first one: equal to it for every value of g. Pick a value, say g = 20, and evaluate both:

\begin{aligned} 15(17) + 4(17) + 30 &= 255 + 68 + 30 \\ &= 353 \\ 19(20) - 27 &= 380 - 27 \\ &= 353 \end{aligned}

They agree. If they had not, there would be a slip somewhere, and the check would have found it without saying where. Choose the test value with some care. At g = 0 every term with a g in it vanishes, so a wrong coefficient of g goes unseen. At g = 1 a coefficient and a constant can trade places unnoticed: 19g - 27 and -27g + 19 both give -8 there. A value such as 2, 3 or 20 avoids both.

Another route for this one. Both brackets here are the same quantity, g - 3. Fifteen of it plus four of it is nineteen of it, exactly as 15x + 4x = 19x, so the bill is also 19(g - 3) + 30, which opens to 19g - 57 + 30 = 19g - 27. That is the distributive property read backwards, which is factoring. It only works when the brackets match; opening first always works.

Part 2: signs in front of a bracket, and fractions

A negative number in front. In 7(a + 2) - 3(a - 5) the minus sign belongs to the 3: subtracting 3(a - 5) is adding (-3)(a - 5), and, as in The distributive property with variables, the -3 multiplies every term inside, so (-3)(-5) = +15.

\begin{aligned} 7(a + 2) - 3(a - 5) &= 7a + 14 - 3a + 15 \\ &= 4a + 29 \end{aligned}

Subtracting a whole bracket. With no number written, the minus sign in 10y - (3y - 8) is a multiplier of -1, so every term inside changes sign, -(3y - 8) = -3y + 8, not just the first:

\begin{aligned} 10y - (3y - 8) &= 10y - 3y + 8 \\ &= 7y + 8 \end{aligned}

A plus sign in front changes nothing, so +(3y - 8) is simply 3y - 8.

Fraction and decimal coefficients. The method does not change. The number in front multiplies each term, \displaystyle \frac{1}{2}(4x + 10) = 2x + 5, and decimals multiply the same way. The only extra work comes when like terms have unlike fractions as coefficients, and then they are added over a common denominator:

\begin{aligned} &\tfrac{1}{2}(x + 4) + \tfrac{1}{3}(x - 3) \\ &= \tfrac{1}{2}x + 2 + \tfrac{1}{3}x - 1 \\ &= \tfrac{3}{6}x + \tfrac{2}{6}x + 1 \\ &= \tfrac{5}{6}x + 1 \end{aligned}

Worked examples

Examples 1 to 3 belong to Part 1, and Examples 4 to 7 to Part 2.

Example 1

One bracket and some loose terms

Simplify (a) 4(2x + 5) + 3x - 7 \qquad (b) 9 + 2(3y - 4) - y

Solution

(a) Distribute the 4 over both terms in the bracket, 4 \cdot 2x = 8x and 4 \cdot 5 = 20, and copy the loose terms unchanged:

\begin{aligned} &4(2x + 5) + 3x - 7 \\ &= 8x + 20 + 3x - 7 \end{aligned}

The like terms are 8x and 3x, and the constants 20 and -7. Gathering them:

8x + 3x + 20 - 7 = 11x + 13

(b) The bracket sits in the middle, which makes no difference. Distribute the 2: 2 \cdot 3y = 6y and 2 \cdot (-4) = -8. The -y at the end is -1y.

\begin{aligned} 9 + 2(3y - 4) - y &= 9 + 6y - 8 - y \\ &= 5y + 1 \end{aligned}

The y terms give 6y - y = 5y, the constants 9 - 8 = 1, and the usual order puts the variable term first.

Example 2

Two bracketed parts

Simplify 3(2a + 4) + 5(a - 1), and check the result at a = 2.

Solution

Open each bracket with its own multiplier. The 3 reaches only the first bracket and the 5 only the second:

\begin{aligned} &3(2a + 4) + 5(a - 1) \\ &= 6a + 12 + 5a - 5 \end{aligned}

Gather the a terms and the constants:

6a + 5a + 12 - 5 = 11a + 7

Check at a = 2. The original gives 3(4 + 4) + 5(2 - 1), that is 24 + 5 = 29, and the result gives 11(2) + 7 = 29. They agree.

Example 3

Two letters, and a check that tests both

Simplify 2(3m + 2n) + 5(m - n + 2), and check the result.

Solution

The second bracket has three terms, and the 5 multiplies all three:

\begin{aligned} &2(3m + 2n) + 5(m - n + 2) \\ &= 6m + 4n + 5m - 5n + 10 \end{aligned}

There are three kinds of term now: m terms, n terms and constants. Each kind is collected on its own, and m and n never combine with each other:

\begin{aligned} &(6m + 5m) + (4n - 5n) + 10 \\ &= 11m - n + 10 \end{aligned}

The n terms give 4n - 5n = -1n, which is written -n.

Check. With two letters, give each its own value, and make the two values different so that an m term mistaken for an n term would show. Take m = 3 and n = 2. The original is 2(9 + 4) + 5(3 - 2 + 2), which is 26 + 15 = 41, and the result is 11(3) - 2 + 10 = 41.

Example 4

A negative number in front

Simplify (a) 5(2x - 3) - 4(x - 6) \qquad (b) -3(2 - w) + 7w

Solution

(a) Read the second part as adding (-4)(x - 6). The -4 multiplies both terms inside: (-4)(x) = -4x, and (-4)(-6) = +24, since the product of two negatives is positive.

\begin{aligned} 5(2x - 3) - 4(x - 6) &= 10x - 15 - 4x + 24 \\ &= 6x + 9 \end{aligned}

Check at x = 2: the original is 5(1) - 4(-4) = 5 + 16 = 21, and 6(2) + 9 = 21.

(b) Here the bracket is first and its multiplier is -3. Inside, the constant comes before the variable, which changes nothing: (-3)(2) = -6 and (-3)(-w) = 3w.

\begin{aligned} -3(2 - w) + 7w &= -6 + 3w + 7w \\ &= 10w - 6 \end{aligned}

Check at w = 2: the bracket is 0, so the original is 0 + 14 = 14, and 10(2) - 6 = 14.

Example 5

Subtracting a whole bracket

Simplify (a) 6y - (2y - 9) \qquad (b) (5p + 3) - (8p + 3)

Solution

(a) The minus sign in front of the bracket is a multiplier of -1, so every term inside changes sign: -(2y - 9) = -2y + 9.

\begin{aligned} 6y - (2y - 9) &= 6y - 2y + 9 \\ &= 4y + 9 \end{aligned}

(b) The first bracket has a plus sign in front (none is written, and none is needed), so it drops away unchanged. The second is subtracted, so both of its terms change sign:

\begin{aligned} (5p + 3) - (8p + 3) &= 5p + 3 - 8p - 3 \\ &= -3p \end{aligned}

The constants cancel, 3 - 3 = 0, and nothing is written for them. Check at p = 2: 13 - 19 = -6, and -3(2) = -6.

Example 6

Fraction and decimal coefficients

Simplify (a) \displaystyle \frac{1}{2}(6x - 4) + \frac{1}{3}(3x + 9) \qquad (b) 0.4(5t + 2) - 0.3(2t - 6)

Solution

(a) Half of each term in the first bracket, and a third of each term in the second:

\begin{aligned} &\tfrac{1}{2}(6x - 4) + \tfrac{1}{3}(3x + 9) \\ &= 3x - 2 + x + 3 \\ &= 4x + 1 \end{aligned}

Every fraction was gone after the first step, because 6, 4, 3 and 9 divide exactly. That is a convenience, not a requirement; the procedure showed a case where fractions remain. Check at x = 2: \displaystyle \frac{1}{2}(8) + \frac{1}{3}(15) is 4 + 5 = 9, and 4(2) + 1 = 9.

(b) The multiplier of the second bracket is -0.3. It multiplies 2t to give -0.6t, and -6 to give +1.8:

\begin{aligned} 0.4(5t + 2) - 0.3(2t - 6) &= 2t + 0.8 - 0.6t + 1.8 \\ &= 1.4t + 2.6 \end{aligned}

Check at t = 1: 0.4(7) - 0.3(-4) is 2.8 + 1.2 = 4, and 1.4 + 2.6 = 4.

Example 7

The perimeter of a shape made from two rectangles

Rectangle A is 3x + 4 long and x + 1 tall. Rectangle B is x wide and x + 3 tall, and it stands on top of A at its left end, so the two share an edge of length x. Find the perimeter of the L-shaped figure they make, as simply as possible.

Figure 2The shape in Example 7, drawn for x = 2, when A is 10 by 3 and B is 2 by 5.

Solution

Add the two rectangles’ perimeters, then take away the shared edge. That edge was counted once in each perimeter, but it lies inside the L, not on its outline, so it is subtracted twice. Each perimeter is twice the length plus twice the width:

\begin{aligned} P &= 2(3x + 4) + 2(x + 1) \\ &\quad + 2x + 2(x + 3) - 2x \\ &= 6x + 8 + 2x + 2 \\ &\quad + 2x + 2x + 6 - 2x \end{aligned}

The x terms give 6x + 2x + 2x + 2x - 2x = 10x, and the constants 8 + 2 + 6 = 16:

P = 10x + 16

Check another way. The outline is made of horizontal and vertical edges only. The horizontal edges along the top add up to the bottom edge, 3x + 4, and the vertical edges on the right add up to the left edge, (x + 1) + (x + 3). So the perimeter is also

\begin{aligned} &2(3x + 4 + x + 1 + x + 3) \\ &= 2(5x + 8) = 10x + 16 \end{aligned}

the same. For x = 2, as drawn, A is 10 by 3 and B is 2 by 5, and walking round the outline gives 10 + 3 + 8 + 5 + 2 + 8 = 36, while 10(2) + 16 = 36.

Boundaries

The number must multiply the bracket. Dividing a bracket by a number distributes, because dividing by 2 is multiplying by \displaystyle \frac{1}{2}: \displaystyle \frac{6x + 4}{2} = 3x + 2. Dividing a number by a bracket does not. \displaystyle \frac{12}{x + 2} is not \displaystyle \frac{12}{x} + \frac{12}{2}; at x = 2 the first is 3 and the second is 12.

A power on a bracket is not a number in front. (x + 3)^2 means (x + 3)(x + 3), a bracket times a bracket, and opening it needs every term of one times every term of the other, which is multiplying two binomials. It is not x^2 + 9: at x = 1 the first is 16 and the second 10.

A sign in front reaches its own bracket and no further. In 8 - (x + 2) + 5x only the x and the 2 change sign; the 5x is outside the bracket and keeps its own. The result is 8 - x - 2 + 5x = 4x + 6.

Only like terms combine. 11m - n + 10 is finished, because m, n and a constant are three different kinds of term. In the same way an x term and an x^2 term stay separate.

What a check by substitution proves. Different results at one value prove a slip. Equal results at one value do not prove the two expressions equivalent: 2x + 1 and 3x are different expressions, yet both give 3 at x = 1. For expressions of the form ax + b, with one letter and no powers, agreeing at two different values does prove them the same (Exercise 24). With more letters, or with powers, one agreement is evidence and no more.

Where this goes next

Solving an equation with brackets starts with exactly this move. In 4(x + 3) - (x - 2) = 20, opening and collecting turns the left side into 3x + 14, and the equation becomes 3x + 14 = 20, so x = 2. That is the work of multi-step equations.

Factoring out a common factor runs the move backwards, as the second route to Gaspard’s bill did. Multiplying two binomials distributes a whole bracket over another one, then combines like terms in the same way, and the products only grow from there. And rewriting an expression to show its meaning is the reason to do all this in the first place: 19g - 27 told Gaspard the cost of each guest in a way 15(g - 3) + 4(g - 3) + 30 did not.

Exercises

Part 1: one bracket and loose terms

Simplify.

5(x + 3) + 2x
6 + 2(4y - 1)

Part 1: two bracketed parts

Simplify.

4(3a - 2) + 2(a + 7)
5(2p + q) + 3(p - 2q)
7(w - 1) + 2(3 - w) + w

Part 1: checking by substituting

Delphine simplified 4(2x + 3) + 5(x - 1) to 13x + 7. Check her answer at x = 2. Is it correct?
Ezekiel simplified 2(x + 6) + 3(2x + 1) to 8x + 9. Check his answer at x = 3, and if it is wrong, find the correct form.

Part 2: a negative number in front

Simplify.

4(x + 5) - 3(x + 2)
6(2k - 1) - 5(k - 3)
-2(4m - 3) + 3(m + 2)

Part 2: subtracting a whole bracket

Simplify.

9y - (4y + 7)
(2r + 11) - (5r - 4)
10 - (3 - 2z) + 4(z - 1)

Part 2: fraction and decimal coefficients

Simplify.

\displaystyle \frac{1}{2}(8x + 6) + \frac{1}{4}(4x - 12)
\displaystyle \frac{2}{3}(9a - 3) - \frac{1}{2}(2a + 10)
0.5(4t + 3) + 0.2(5t - 10)
1.5(2c - 4) - 0.4(5c + 5)

Part 2: totals, orders and perimeters

Cyprian books a museum trip for n students. Each student needs a ticket costing 6 dollars and a bus seat costing 2 dollars, but 5 students drop out, and a booking fee of 25 dollars is added. Write the cost as an expression with brackets, simplify it, and find the cost when n = 30.
Honoria cycles 2x + 3 kilometers on each of 4 days and x - 2 kilometers on each of 3 days. Write her total distance with brackets, simplify it, and find it when x = 5.
Two identical rectangles, each 2x + 1 long and x + 3 wide, are placed end to end with their x + 3 sides together, making one long rectangle. (a) Find the perimeter of one rectangle. (b) Explain why the long rectangle’s perimeter is 2(6x + 8) - 2(x + 3), and simplify it. (c) Check by finding the long rectangle’s length and width directly.
Fenella has 3 pieces of ribbon, each 2x + 9 centimeters long. She uses 4x - 1 centimeters of ribbon in total. Write the length left over with brackets, and simplify it.

Reasoning

Explain why 4 + 3(x + 2) is not equal to 7(x + 2), and find its simplest form.
Simplify 5(2x + 3) + 2(x - 4). Then explain why x = 0 is a poor value at which to check the answer.
Two expressions ax + b and cx + d give equal values at x = p and at x = q, where p \ne q. Prove that a = c and b = d, so they are the same expression.
Show that 2(3x + 1) - 3(2x - 4) has the same value whatever x is, and find that value.

Answers

7x + 15. 5x + 15 + 2x.
8y + 4. 6 + 8y - 2.
14a + 6. 12a - 8 + 2a + 14.
13p - q. 10p + 5q + 3p - 6q, and 5q - 6q = -q.
6w - 1. 7w - 7 + 6 - 2w + w.
Yes. At x = 2 the original is 4(7) + 5(1) = 33, and 13(2) + 7 = 33. Opening the brackets confirms it: 8x + 12 + 5x - 5 = 13x + 7.
Not correct; the simplest form is 8x + 15. At x = 3 the original is 2(9) + 3(7) = 39, but 8(3) + 9 = 33. Opening the brackets gives 2x + 12 + 6x + 3 = 8x + 15, which is 39 at x = 3. The 9 suggests the 2 was never multiplied into the 6.
x + 14. 4x + 20 - 3x - 6.
7k + 9. 12k - 6 - 5k + 15, since (-5)(-3) = 15.
-5m + 12. -8m + 6 + 3m + 6, since (-2)(-3) = 6.
5y - 7. 9y - 4y - 7.
-3r + 15. 2r + 11 - 5r + 4.
6z + 3. 10 - 3 + 2z + 4z - 4.
5x. 4x + 3 + x - 3, and the constants cancel.
5a - 7. 6a - 2 - a - 5.
3t - 0.5. 2t + 1.5 + t - 2.
c - 8. 3c - 6 - 2c - 2.
The cost is 6(n - 5) + 2(n - 5) + 25, which simplifies to 8n - 15, and it is 225 dollars when n = 30. Opening: 6n - 30 + 2n - 10 + 25. Check: 25 students at 8 dollars is 200, plus 25 is 225.
The total is 4(2x + 3) + 3(x - 2), which simplifies to 11x + 6, and 61 kilometers when x = 5. Opening: 8x + 12 + 3x - 6. Check: 4(13) + 3(3) = 52 + 9 = 61.
(a) 6x + 8, from 2(2x + 1) + 2(x + 3), which opens to 4x + 2 + 2x + 6. (b) The two perimeters together are 2(6x + 8), but each counts the joined x + 3 side once, and that side is inside the long rectangle, so it comes off twice: 12x + 16 - 2x - 6 = 10x + 10. (c) The long rectangle is 2(2x + 1) = 4x + 2 long and x + 3 wide, so its perimeter is 2(4x + 2) + 2(x + 3), which opens to 8x + 4 + 2x + 6 and simplifies to 10x + 10, the same.
3(2x + 9) - (4x - 1) = 2x + 28 centimeters. Opening: 6x + 27 - 4x + 1; the -1 becomes +1 because the whole bracket is subtracted.
4 + 3(x + 2) = 3x + 10. The 3 multiplies the bracket and the 4 is added afterwards, by the order of operations, so the 4 and the 3 cannot be added first. The two expressions differ: at x = 1 the first is 4 + 9 = 13 and the second is 21.
12x + 7, from 10x + 15 + 2x - 8. At x = 0 every x term is 0, so any answer ending in + 7 passes, whatever its coefficient of x: 11x + 7 would pass too. A value such as x = 2 tests the coefficient as well.
From ap + b = cp + d and aq + b = cq + d, subtract the second from the first: a(p - q) = c(p - q). Since p \ne q, divide by p - q to get a = c. Then ap + b = ap + d gives b = d.
The value is always 14. Opening: 6x + 2 - 6x + 12, and the x terms cancel, 6x - 6x = 0, leaving 2 + 12 = 14.