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Rearranging equations

Authored by VIrginia Vuong

Mathematics

1st - 12th Grade

CCSS covered

Used 4+ times

Rearranging equations
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8 questions

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1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Select the equation which is not equivalent to the rest.

 ab = cab\ =\ c  

 a = bca\ =\ bc  

 c = abc\ =\ \frac{a}{b}  

 bc = abc\ =\ a  

 b = acb\ =\ \frac{a}{c}  

Tags

CCSS.HSA.CED.A.4

2.

MULTIPLE SELECT QUESTION

1 min • 1 pt

Select all equations which are equivalent to   3 = abc3\ =\ abc  

 a = 3bca\ =\ \frac{3}{bc}  

 b = 3acb\ =\ \frac{3}{ac}  

 c = 3abc\ =\ \frac{3}{ab}  

 b = 3cab\ =\ \frac{3}{ca}  

 3a = bc\frac{3}{a}\ =\ bc  

Tags

CCSS.HSA.CED.A.4

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Select the equation which is not equivalent to the rest.

a = b +cda\ =\ b\ +cd

b = a + cdb\ =\ a\ +\ cd

b = a cdb\ =\ a\ -\ cd

c = a bdc\ =\ \frac{a\ -\ b}{d}

d = abcd\ =\ \frac{a-b}{c}

Tags

CCSS.HSA.CED.A.4

4.

MULTIPLE SELECT QUESTION

1 min • 1 pt

Select all the equations which are equivalent to 3x + 2y = 73x\ +\ 2y\ =\ 7  

 7  3x = 2y7\ -\ 3x\ =\ 2y  

 7  2y = 3x7\ -\ 2y\ =\ 3x  

 7 + 3x = 2y7\ +\ 3x\ =\ 2y  

 7 + 2y = 3x7\ +\ 2y\ =\ 3x  

 y = 7  3x2y\ =\ \frac{7\ -\ 3x}{2}  

Tags

CCSS.HSA.CED.A.4

5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Select the equation which is not equivalent to the rest.

b = 3(c5)b\ =\ 3\left(c-5\right)

c = b3+5c\ =\ \frac{b}{3}+5

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

b = 3c 15b\ =\ 3c\ -\ 15

c = 15b3c\ =\ \frac{15-b}{3}

Tags

CCSS.HSA.CED.A.4

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Select the option which has this equation rearranged with n as the subject: 5m = n95m\ =\ n-9  

 m = n95m\ =\ \frac{n-9}{5}  

 5m  n = 95m\ -\ n\ =\ -9  

 n = 5m + 9n\ =\ 5m\ +\ 9  

 n = 5m  9n\ =\ 5m\ -\ 9  

 n = 5m9n\ =\ \frac{5m}{9}  

Tags

CCSS.HSA.CED.A.4

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

 a = b  cd + ea\ =\ \frac{b\ -\ c}{d\ +\ e}  

(CHALLENGE) Choose the option which has this equation rearranged to have d as the subject:

 d = b  ca  ed\ =\ \frac{b\ -\ c}{a}\ -\ e  

 d = b  ca + ed\ =\ \frac{b\ -\ c}{a\ +\ e}  

 d = b  ca  ed\ =\ \frac{b\ -\ c}{a\ -\ e}  

 d = e + bcad\ =\ e\ +\ \frac{b-c}{a}  

 d = e + a  eb  cd\ =\ e\ +\ \frac{a\ -\ e}{b\ -\ c}  

Tags

CCSS.HSA.CED.A.4

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