Equivalent Expressions: Grade 7 Math Worksheet Answer Key – PDF Download

Every solution below shows each step, so compare your method as well as your final answer. Final answers are in bold. When an answer differs from yours, substitute a number for the variable in both expressions: equivalent expressions always give the same value.

Grade 7Common Core: 7.EE.A.1 · 7.EE.A.2

Problem 1: Combine like terms

Both terms have the variable $x$, so add the coefficients: $6+9=15$.

$6x+9x=$$15x$

The coefficients add: 6 + 9 = 15.

Problem 2: Algebra tiles

Group the x tiles: $3x + 4x = 7x$. Group the unit tiles: $2 + 5 = 7$.

$3x + 2 + 4x + 5 = 7x + 7$

Group the like tiles: 7 x tiles and 7 units.

Problem 3: Collect terms and constants

Group the like terms: $(7y+2y)+(4-9)$.

$7y+2y=9y$ and $4-9=-5$.

$7y+4+2y-9=$$9y-5$

Check with $y=1$: $7+4+2-9=4$ and $9-5=4$.

Problem 4: Expand with the distributive property

Multiply 5 by each term inside the parentheses: $5\cdot x=5x$ and $5\cdot 6=30$.

$5(x+6)=$$5x+30$

The two areas are 5x and 30.

Problem 5: Area model for expanding

Left area: $7 \times 3x = 21x$. Right area: $7 \times 2 = 14$.

Add the areas: $7(3x + 2) = 21x + 14$

The areas are 21x and 14.

Problem 6: Test equivalence with a table

Substitute each value. For $x = 1$: $2(1 + 3) = 8$, $2(1) + 6 = 8$, and $2(1) + 3 = 5$. The other columns follow the same way.

The first two rows match.

$2(x + 3)$ and $2x + 6$ give the same value in every column. By the distributive property, $2(x + 3) = 2x + 6$, so they are equivalent.

$2x + 3$ is always 3 less, so it is not equivalent to the other two.

Problem 7: Expand with a negative factor

$-3\cdot 2x=-6x$ and $-3\cdot(-4)=+12$.

$-3(2x-4)=$$-6x+12$

Check with $x=1$: $-3(-2)=6$ and $-6+12=6$.

Problem 8: Expand and combine

$6 \times 2y = 12y$ and $6 \times (-3) = -18$. $12y – 18$
$-2 \times 5 = -10$ and $-2 \times (-3a) = 6a$. $-10 + 6a$
$4(x + 2) = 4x + 8$ and $3(x – 1) = 3x – 3$. Add: $4x + 3x = 7x$ and $8 – 3 = 5$. $7x + 5$

Problem 9: Factor out the greatest common factor

The common factors of 12 and 18 are 1, 2, 3 and 6; the greatest is $6$.

$12x: 6=2x$ and $18: 6=3$.

$12x+18=$$6(2x+3)$

Height 6, widths 2x and 3.

Problem 10: Factor out a common factor

GCF of 15 and 25 is 5. $5(3a + 5)$
GCF of 8 and 12 is 4. $4(2x – 3)$
GCF of 18 and 54 is 18. $18(m + 3)$
Factor out $-3$: $-6y = -3 \times 2y$ and $-9 = -3 \times 3$. $-3(2y + 3)$

Problem 11: Subtract an expression

Subtracting changes the sign of each term in the second expression: $3x+8-x+5$.

$3x-x=2x$ and $8+5=13$.

$(3x+8)-(x-5)=$$2x+13$

Check with $x=1$: $11-(-4)=15$ and $2+13=15$.

Problem 12: Subtract and simplify

Distribute the minus sign: $7x + 3 – 2x – 9 = 5x – 6$. $5x – 6$
$2m – 5 – m + 8 = m + 3$. $m + 3$
$10 – 3a – 4 + a = 6 – 2a$. $6 – 2a$

Problem 13: Rational coefficients

Distribute: $\frac{2}{3}\cdot 9x=6x$ and $\frac{2}{3}\cdot(-6)=-4$, so the expression is $6x-4+x$.

Combine $6x+x=7x$.

$\frac{2}{3}(9x-6)+x=$$7x-4$

Check with $x=3$: $\frac{2}{3}(21)+3=17$ and $21-4=17$.

Problem 14: Decimal and fraction coefficients

$0.5 \times 6x = 3x$ and $0.5 \times 10 = 5$, so $3x + 5 + 1.5x = 4.5x + 5$. $4.5x + 5$
$\frac{1}{3} \times 12x = 4x$ and $\frac{1}{3} \times 9 = 3$, so $4x + 3 – x = 3x + 3$. $3x + 3$
$\frac{3}{4} \times 8 = 6$ and $\frac{3}{4} \times (-4x) = -3x$, so $6 – 3x + x = 6 – 2x$. $6 – 2x$

Problem 15: Sales tax at the school store

(a) $p+0.08p=1p+0.08p=$$1.08p$.

(b) $1.08(25)=27$, so the total is $\$27.00$.

The total is 1.08 times the price.

Problem 16: Discount, then tax

(a) $p – 0.15p = 0.85p$.

(b) Adding 8% tax means multiplying by 1.08: $1.08 \times 0.85p = 0.918p$.

(d) $0.918 \times 100 = 91.8$, so the regular price is $100.

$0.85p$; $0.918p$; $55.08; $100.

Problem 17: Perimeter of a garden bed

(a) The perimeter is twice the length plus twice the width: $2(3x+2)+2(x+5)=6x+4+2x+10=$$8x+14$ feet.

(b) $8(3)+14=38$. 38 feet.

Check: the sides are $3(3)+2=11$ ft and $3+5=8$ ft, and $2(11)+2(8)=38$.

Perimeter: 8x + 14.

Problem 18: Perimeter of a triangular bed

(a) $(2x + 3) + (x + 7) + (3x – 1) = (2x + x + 3x) + (3 + 7 – 1) = 6x + 9$ feet.

(b) $6(4) + 9 = 33$ ft. The sides are $2(4) + 3 = 11$, $4 + 7 = 11$, and $3(4) – 1 = 11$, so the bed is an equilateral triangle.

(c) Set $6x + 9 = 45$. Then $6x = 36$ and $x = 6$. The sides are 15 ft, 13 ft, and 17 ft, and $15 + 13 + 17 = 45$.

$6x + 9$; 33 ft (equilateral); $x = 6$ with sides 15, 13, 17 ft.

Problem 19: Three ways to write a gym bill

(a) Maya: $4(c + 3) = 4c + 12$. Noah: $3(c + 4) + c = 3c + 12 + c = 4c + 12$.

(b) Ethan’s $12 + 4c$ is $4c + 12$ in a different order. All three equal $4c + 12$, so they are equivalent.

Equivalent; $48 for 9 classes; $32 for 5 classes.

Problem 20: Two ways to price a jersey

(a) Maya: $p-0.20p-5=0.80p-5$. Ethan: $0.80(p-5)=0.80p-4$.

(b) The expressions are not equivalent: $0.8p-5$ and $0.8p-4$ differ by $1$ for every price.

(c) Maya: $0.8(40)-5=32-5=$$\$27$. Ethan: $0.8(40)-4=32-4=28$, which is $\$28$. Check Ethan directly: $0.8(35)=28$. Maya pays $\$1$ less.

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