Solve for y: 3 − 4 y = 18 \frac{3}{-4}y\ =\ 18 − 4 3 y = 1 8

y = − 27 2 y=-\frac{27}{2} y = − 2 2 7

b + 4 5 = 9 10 b\ +\ \frac{4}{5}=\ \frac{9}{10} b + 5 4 = 1 0 9

b = 1 10 b\ =\frac{1}{10} b = 1 0 1

b = 13 15 b\ =\ \frac{13}{15} b = 1 5 1 3

b = 5 5 b\ =\frac{5}{5} b = 5 5

b = 13 5 b\ =\frac{13}{5} b = 5 1 3

g + 9 10 = 7 10 g\ +\ \frac{9}{10}=\frac{7}{10} g + 1 0 9 = 1 0 7

g = − 1 5 g\ =-\frac{1}{5} g = − 5 1

g = 1 5 g\ =\frac{1}{5} g = 5 1

g = 16 10 g\ =\frac{16}{10} g = 1 0 1 6

g = 16 20 g\ =\frac{16}{20} g = 2 0 1 6

m + 3 4 = 1 4 m\ +\frac{3}{4}\ =\frac{1}{4} m + 4 3 = 4 1

m = − 1 2 m=-\frac{1}{2} m = − 2 1

m = 31 44 m=\frac{31}{44} m = 4 4 3 1

a − 1 8 = 5 8 a-\frac{1}{8}=\frac{5}{8} a − 8 1 = 8 5

a = 5 64 a=\frac{5}{64} a = 6 4 5

t − 2 3 = 4 9 t\ -\frac{2}{3}=\frac{4}{9} t − 3 2 = 9 4

t = 1 1 9 t=1\ \frac{1}{9} t = 1 9 1

t = 1 6 12 t=1\ \frac{6}{12} t = 1 1 2 6

c − 9 10 = 1 3 c\ -\frac{9}{10}=\frac{1}{3} c − 1 0 9 = 3 1

c = 1 7 30 c\ =1\ \frac{7}{30} c = 1 3 0 7

c = 1 10 13 c=1\ \frac{10}{13} c = 1 1 3 1 0

c = 1 1 3 c=1\ \frac{1}{3} c = 1 3 1

c = 10 30 c=\frac{10}{30} c = 3 0 1 0

A carpenter used 3 3 16 3\ \frac{3}{16} 3 1 6 3 lb of nails for a job. After the job was over, the remaining nails weighted 1 1 16 1\ \frac{1}{16} 1 1 6 1 lb. How many pounds of nails did the carpenter have at the beginning of the job?

3 48 16 3\ \frac{48}{16} 3 1 6 4 8

3 1 2 n = 28 3\ \frac{1}{2}n\ =\ 28 3 2 1 n = 2 8

n = 21 2 n=\frac{21}{2} n = 2 2 1

n = 35 2 n=\frac{35}{2} n = 2 3 5

6 p = 5 8 6p\ =\ \frac{5}{8} 6 p = 8 5

p = 5 48 p\ =\ \frac{5}{48} p = 4 8 5

p = 30 8 p\ =\ \frac{30}{8} p = 8 3 0

p = 11 8 p\ =\frac{11}{8} p = 8 1 1

p = − 1 8 p\ =\ -\frac{1}{8} p = − 8 1

1 3 y = 2 \frac{1}{3}y\ =\ 2 3 1 y = 2

y = 1 6 y\ =\frac{1}{6} y = 6 1

y = 2 3 y\ =\ \frac{2}{3} y = 3 2

− 1 1 10 t = 1 15 -1\ \frac{1}{10}t\ =\ \frac{1}{15} − 1 1 0 1 t = 1 5 1

t = − 2 33 t\ =\ -\frac{2}{33} t = − 3 3 2

t = 10 15 t=\frac{10}{15} t = 1 5 1 0

t = 11 150 t=\frac{11}{150} t = 1 5 0 1 1

SOLVING EQUATIONS WITH FRACTIONS

Authored by Abiel Guajardo

Mathematics

7th - 8th Grade

CCSS covered

Used 304+ times

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20 questions

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MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Solve for y:
$\frac{3}{-4}y\ =\ 18$

$y=24$

$y=-24$

$y=-\frac{27}{2}$

$y=-72$

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

x + ⅚ = ⅔

x = -⅙

x = ⅙

x = 1½

x = -1½

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

x + 1¹/₂= 5³/₄

x = 4 ¹/₄

x = 4 ⁵/₄

x = 4 ²/₂

x = 5

Tags

CCSS.6.EE.B.7

CCSS.6.EE.B.5

CCSS.6.EE.B.6

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

x - ⁵/₆ = 10

x = 10 ⁵/₆

x = 9 ¹/₆

x = 9 ⁵/₆

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

⅔m = -4

m = 2 ⅔

m = -6

m = 6

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

d - 3/7 = 5/7

d = 8/7

d = 6/7

d = 4/7

d = 3/7

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

3y = 6/7

y = 1/7

y = 2/7

y = 3/7

y = 4/7

Tags

CCSS.6.EE.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$b\ +\ \frac{4}{5}=\ \frac{9}{10}$

$b\ =\frac{1}{10}$

$b\ =\ \frac{13}{15}$

$b\ =\frac{5}{5}$

$b\ =\frac{13}{5}$

Tags

CCSS.5.NF.A.1

CCSS.5.NF.A.2

10.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$g\ +\ \frac{9}{10}=\frac{7}{10}$

$g\ =-\frac{1}{5}$

$g\ =\frac{1}{5}$

$g\ =\frac{16}{10}$

$g\ =\frac{16}{20}$

Tags

CCSS.6.EE.B.7

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SOLVING EQUATIONS WITH FRACTIONS

Solve for y: 3−4y = 18\frac{3}{-4}y\ =\ 18−43​y = 18

x + ⅚ = ⅔

x + 11/2 = 53/4

x - 5/6 = 10

⅔m = -4

d - 3/7 = 5/7

3y = 6/7

h - 3/8 = -11/24

b + 45= 910b\ +\ \frac{4}{5}=\ \frac{9}{10}b + 54​= 109​

g + 910=710g\ +\ \frac{9}{10}=\frac{7}{10}g + 109​=107​

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Solve for y:
$\frac{3}{-4}y\ =\ 18$

x + 1¹/₂= 5³/₄

x - ⁵/₆ = 10

$b\ +\ \frac{4}{5}=\ \frac{9}{10}$

$g\ +\ \frac{9}{10}=\frac{7}{10}$