Solving Equations (LCD Method)

Authored by Ryder Weilbacher

Mathematics

9th - 12th Grade

CCSS covered

Used 13+ times

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12 questions

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{5}{6}+3m=-\frac{2}{3}$

m=-1/2

m=1/2

m=-9

m=9

Tags

CCSS.7.EE.B.4A

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{2}{9}g+\frac{2}{3}=\frac{1}{2}g-\frac{1}{2}$

g = -3

g = 21/5

g = -21/5

g = 3

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{1}{2}\left(5-3m\right)=-\frac{2}{3}$

m=19/9

m=9/19

m=-9

m=9

Tags

CCSS.7.EE.B.4A

MATH RESPONSE QUESTION

2 mins • 1 pt

Solve

$3$

Mathematical Equivalence

Tags

CCSS.8.EE.C.7B

MATH RESPONSE QUESTION

2 mins • 1 pt

$\frac{3}{4}-\frac{5}{4}m=\frac{113}{24}$

$-\frac{19}{6}$

Mathematical Equivalence

Tags

CCSS.7.EE.B.4A

MATH RESPONSE QUESTION

2 mins • 1 pt

Solve

$20$

Mathematical Equivalence

Tags

CCSS.8.EE.C.7B

MATH RESPONSE QUESTION

2 mins • 1 pt

$\frac{3}{2}\left(\frac{7}{3}n+1\right)=\frac{3}{2}$

$0$

Mathematical Equivalence

Tags

CCSS.7.EE.B.4A

MATH RESPONSE QUESTION

2 mins • 2 pts

Solve

$-48$

Mathematical Equivalence

Tags

CCSS.8.EE.C.7B

MATH RESPONSE QUESTION

2 mins • 1 pt

$2b+\frac{9}{5}=-\frac{11}{5}$

$-2$

Mathematical Equivalence

Tags

CCSS.7.EE.B.4A

10.

MATH RESPONSE QUESTION

2 mins • 1 pt

Solve

$\frac{2}{3}$

Mathematical Equivalence

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Solving Equations (LCD Method)

56+3m=−23\frac{5}{6}+3m=-\frac{2}{3}65​+3m=−32​

29g+23=12g−12\frac{2}{9}g+\frac{2}{3}=\frac{1}{2}g-\frac{1}{2}92​g+32​=21​g−21​

12(5−3m)=−23\frac{1}{2}\left(5-3m\right)=-\frac{2}{3}21​(5−3m)=−32​

Solve

34−54m=11324\frac{3}{4}-\frac{5}{4}m=\frac{113}{24}43​−45​m=24113​

Solve

32(73n+1)=32\frac{3}{2}\left(\frac{7}{3}n+1\right)=\frac{3}{2}23​(37​n+1)=23​

Solve

2b+95=−1152b+\frac{9}{5}=-\frac{11}{5}2b+59​=−511​

Solve

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$\frac{5}{6}+3m=-\frac{2}{3}$

$\frac{2}{9}g+\frac{2}{3}=\frac{1}{2}g-\frac{1}{2}$

$\frac{1}{2}\left(5-3m\right)=-\frac{2}{3}$

$\frac{3}{4}-\frac{5}{4}m=\frac{113}{24}$

$\frac{3}{2}\left(\frac{7}{3}n+1\right)=\frac{3}{2}$

$2b+\frac{9}{5}=-\frac{11}{5}$