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Solving Literal Equations

Solving Literal Equations

Assessment

Presentation

•

Mathematics

•

11th Grade

•

Practice Problem

•

Medium

•
CCSS
HSA.CED.A.4, HSA.REI.A.1

Standards-aligned

Created by

Malaka Mallery

Used 2+ times

FREE Resource

4 Slides • 8 Questions

1

Warm up

2

Reorder

Justify the steps in solving the following equation:  25x+6=10\frac{2}{5}x+6=10
Reorder the following

subtract 6 from both sides

multiply both sides by 5

divide both sides by 2

1
2
3

3

Vocabulary

4

Categorize

Options (6)

d=ad=a  

A=12bhA=\frac{1}{2}bh  

m+n=3pm+n=3p  

−4x=20-4x=20  

10=13(y−6)10=\frac{1}{3}\left(y-6\right)  

n+5=2n−14n+5=2n-14  

Which are examples and non-examples of literal equations.

Organize these options into the right categories

Example
Non-example

5

Your Turn

6

Multiple Choice

Solve for g.

s=12gt2s=\frac{1}{2}gt^2

1

g=2st2g=\frac{2s}{t^2}

2

g=st22g=\frac{st^2}{2}

3

g=2st2g=2st^2

4

g=12t2g=\frac{1}{2}t^2

7

Multiple Choice

Solve for R.

Ra−5=b\frac{R}{a}-5=b

1

R=a(b+5)R=a\left(b+5\right)

2

R=ba−5R=\frac{b}{a}-5

3

R=ab+5R=ab+5

4

R=ab+5R=\frac{a}{b}+5

8

Multiple Choice

Solve for the indicated variable.

If there is $1500 invested is in a bank account after 4 years simple interest, the A is the amount of money in the account in 4 years is given by A = 1500(1 + r(4)). 

What is r in terms of A?

1

r=A1500−14r=\frac{\frac{A}{1500}-1}{4}

2

r=4(A−1500)r=\text{4}(A-1500)

3

r=A +15004r=\frac{A\ +1500}{4}

4

r=A−15004r=\frac{A-1500}{4}

9

Exit Ticket

10

Drag and Drop

Solve for x:

2/5 (x + 1) = g



x = ​
Drag these tiles and drop them in the correct blank above

11

Drag and Drop

Solve for the indicated variable.

A = 1/2 h (b1 + b2) for h.

h = ​ ​
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12

Drag and Drop

If $2000 is in a bank account earning 15% simple interest, the dollar value of the account A in t years is given by A = 2000(1 + 0.15t).What is t in terms of A?

t =​
Drag these tiles and drop them in the correct blank above
pattern-tertiary
Warm up

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