M
Math Coach Amy
Amy Ferguson Moncure

Algebra 1 · Unit 1

Literal Equations

Rearrange a formula or equation to isolate a chosen variable, using the properties of equality.

Jump to section:

Key Ideas One Step Two Steps y in terms of x Factor Super Challenge
What to Remember

One Step example

Solve d = rt for r.

d = rt

Divide both sides by t.

dt = rtt

dt = rtt

r = dt

Answer: r = dt

Two Steps example

Solve A = 12bh for b.

A = 12bh

Multiply both sides by 2.

2A = bh

Divide both sides by h.

b = 2Ah

Answer: b = 2Ah

y in terms of x example

What is y in terms of x? 3x + y = 12

y in terms of x means solve for y.

x will be part of the answer.

3x + y = 12

Subtract 3x from both sides.

y = 12 − 3x

Answer: y = 12 − 3x

Factor example

Solve A = 3πrh + 6πr² for h.

A = 3πrh + 6πr²

Subtract 6πr² from both sides.

A − 6πr² = 3πrh

Divide both sides by 3πr.

h = A − 6πr²3πr

Answer: h = A − 6πr²3πr

Level 1

One Step

One step. Isolate the letter.

Example
Solve d = rt for r.

d = rt

Divide both sides by t.

dt = rtt

dt = rtt

r = dt

Answer: r = dt

1
Solve P = 4s for s.
Show Answer

P = 4s

Divide both sides by 4.

P4 = 4s4

P4 = 4s4

s = P4

Answer: s = P4

2
Solve I = Prt for t.
Show Answer

I = Prt

Divide both sides by Pr.

IPr = PrtPr

IPr = PrtPr

t = IPr

Answer: t = IPr

3
Solve y = mx + b for b.
Show Answer

y = mx + b

Subtract mx from both sides.

y − mx = mx + b − mx

y − mx = mx + b − mx

b = y − mx

Answer: b = y − mx

4
Solve A = ℓw for ℓ.
Show Answer

A = ℓw

Divide both sides by w.

Aw = ℓww

Aw = ℓww

ℓ = Aw

Answer: ℓ = Aw

5
Solve C = πd for d.
Show Answer

C = πd

Divide both sides by π.

Cπ = πdπ

Cπ = πdπ

d = Cπ

Answer: d = Cπ

6
Solve C = 2πr for r.
Show Answer

C = 2πr

Divide both sides by 2π.

C2π = 2πr2π

C2π = 2πr2π

r = C2π

Answer: r = C2π

7
Solve V = ℓwh for h.
Show Answer

V = ℓwh

Divide both sides by ℓw.

Vℓw = ℓwhℓw

Vℓw = ℓwhℓw

h = Vℓw

Answer: h = Vℓw

Level 2

Two Steps

Two steps.

Example
Solve A = 12bh for b.

A = 12bh

Multiply both sides by 2.

2A = bh

Divide both sides by h.

b = 2Ah

Answer: b = 2Ah

1
Solve y = mx + b for x.
Show Answer

y = mx + b

Subtract b from both sides.

y − b = mx

Divide both sides by m.

x = y − bm

Answer: x = y − bm

2
Solve P = 2ℓ + 2w for w.
Show Answer

P = 2ℓ + 2w

Subtract 2ℓ from both sides.

P − 2ℓ = 2w

Divide both sides by 2.

w = P − 2ℓ2

Answer: w = P − 2ℓ2

3
Solve A = p + q2 for q.
Show Answer

A = p + q2

Multiply both sides by 2.

2A = p + q

Subtract p from both sides.

2A − p = q

q = 2A − p

Answer: q = 2A − p

4
Solve ax + b = c for x.
Show Answer

ax + b = c

Subtract b from both sides.

ax = c − b

Divide both sides by a.

x = c − ba

Answer: x = c − ba

Level 2

y in terms of x

y in terms of x means solve for y. x stays in the answer.

Example
What is y in terms of x? 3x + y = 12

y in terms of x means solve for y.

x will be part of the answer.

3x + y = 12

Subtract 3x from both sides.

y = 12 − 3x

Answer: y = 12 − 3x

1
If 7x − 3y = 21, what is y in terms of x?
Show Answer

7x − 3y = 21

Subtract 7x from both sides.

−3y = 21 − 7x

Divide both sides by −3.

y = 7x − 213

or y = 73x − 7

Answer: y = 7x − 213

2
The volume of a rectangular pyramid is V = 13ℓwh. What is w in terms of V, ℓ, and h?
Show Answer

V = 13ℓwh

Multiply both sides by 3.

3V = ℓwh

Divide both sides by ℓh.

w = 3Vℓh

Answer: w = 3Vℓh

3
If 4x − 3y = 18, what is y in terms of x?
Show Answer

4x − 3y = 18

Subtract 4x from both sides.

−3y = 18 − 4x

Divide both sides by −3.

y = 4x − 183

or y = 43x − 6

Answer: y = 4x − 183

Level 3

Factor

Isolate the term that has the variable, or factor that variable from every term.

Example
Solve A = 3πrh + 6πr² for h.

A = 3πrh + 6πr²

Subtract 6πr² from both sides.

A − 6πr² = 3πrh

Divide both sides by 3πr.

h = A − 6πr²3πr

Answer: h = A − 6πr²3πr

1
A cylinder’s surface area is A = 2πrh + 2πr². Solve for h.
Show Answer

A = 2πrh + 2πr²

Subtract 2πr² from both sides.

A − 2πr² = 2πrh

Divide both sides by 2πr.

h = A − 2πr²2πr

Answer: h = A − 2πr²2πr

2
Solve S = 5πr² + 10πrh for π.
Show Answer

S = 5πr² + 10πrh

Factor π from both terms.

S = π(5r² + 10rh)

Divide both sides by 5r² + 10rh.

S5r² + 10rh = π(5r² + 10rh)5r² + 10rh

S5r² + 10rh = π

π = S5r² + 10rh

Answer: π = S5r² + 10rh

3
Solve A = 6xy² + 9xyz for x.
Show Answer

A = 6xy² + 9xyz

Factor x from both terms.

A = x(6y² + 9yz)

Divide both sides by 6y² + 9yz.

A6y² + 9yz = x(6y² + 9yz)6y² + 9yz

A6y² + 9yz = x

x = A6y² + 9yz

Answer: x = A6y² + 9yz

4
Solve B = 4mn² − 10mnp for m.
Show Answer

B = 4mn² − 10mnp

Factor m from both terms.

B = m(4n² − 10np)

Divide both sides by 4n² − 10np.

B4n² − 10np = m(4n² − 10np)4n² − 10np

B4n² − 10np = m

m = B4n² − 10np

Answer: m = B4n² − 10np

Super Challenge

Super Challenge

1
If 5n + 8 = kn − 4, what is n in terms of k?
Show Answer

5n + 8 = kn − 4

Subtract kn from both sides.

5n − kn + 8 = −4

Subtract 8 from both sides.

5n − kn = −12

Factor n from both terms.

n(5 − k) = −12

Divide both sides by 5 − k.

n = −125 − k

or n = 12k − 5

Answer: n = 12k − 5

2
If 3(2x + y) = 4x − 5, what is y in terms of x?
Show Answer

3(2x + y) = 4x − 5

Distribute.

6x + 3y = 4x − 5

Subtract 6x from both sides.

3y = 4x − 6x − 5

3y = −2x − 5

Divide both sides by 3.

y = −2x − 53

Answer: y = −2x − 53

3
Solve 5a = 5b + 5c − 5d for c.
Show Answer

5a = 5b + 5c − 5d

Subtract 5b from both sides.

5a − 5b = 5c − 5d

Add 5d to both sides.

5a − 5b + 5d = 5c

Divide both sides by 5.

1a − 1b + 1d = 1c

The common denominator is abd.

bdabd − adabd + ababd

= 1c

bd − ad + ababd = 1c

Multiply both sides by c.

cbd − ad + ababd = 1

Multiply both sides by abd.

c(bd − ad + ab) = abd

Divide both sides by bd − ad + ab.

c = abdbd − ad + ab

Answer: c = abdbd − ad + ab

4
Solve 3a = 2b − 6c + 1d for b.
Show Answer

3a = 2b − 6c + 1d

Add 6c to both sides.

3a + 6c = 2b + 1d

Subtract 1d from both sides.

3a + 6c − 1d = 2b

The common denominator is acd.

3cdacd + 6adacd − acacd

= 2b

3cd + 6ad − acacd = 2b

Multiply both sides by b.

b3cd + 6ad − acacd = 2

Multiply both sides by acd.

b(3cd + 6ad − ac) = 2acd

Divide both sides by 3cd + 6ad − ac.

b = 2acd3cd + 6ad − ac

Answer: b = 2acd3cd + 6ad − ac

5
Solve A = B − 34(12 − P) for P.
Show Answer

A = B − 34(12 − P)

Distribute.

−34(12 − P)

= −34 · 12 + 34P

34 · 12 = 9

−34(12 − P) = −9 + 34P

A = B − 9 + 34P

Subtract B from both sides.

A − B = −9 + 34P

Add 9 to both sides.

A − B + 9 = 34P

Multiply both sides by 43.

43(A − B + 9) = P

43 · 9 = 12

P = 43(A − B) + 12

P = 12 + 43(A − B)

Answer: P = 12 + 43(A − B)

6
Solve 50A − 5B = −10C for A.
Show Answer

50A − 5B = −10C

Divide every term by 5.

10A − 1B = −2C

Add 1B to both sides.

10A = 1B − 2C

The common denominator is BC.

1B = CBC

−2C = −2BBC

10A = C − 2BBC

Multiply both sides by A.

10 = AC − 2BBC

Multiply both sides by BC.

10BC = A(C − 2B)

Divide both sides by C − 2B.

A = 10BCC − 2B

Answer: A = 10BCC − 2B

7
Solve r = square root of Aπ for A.
Show Answer

r = square root of Aπ

A is under the square root. Square both sides.

r² = Aπ

Multiply both sides by π.

πr² = A

A = πr²

Answer: A = πr²

8
Solve v = square root of 2gh for h.
Show Answer

v = square root of 2gh

h is under the square root. Square both sides.

v² = 2gh

Divide both sides by 2g.

v²2g = h

h = v²2g

Answer: h = v²2g

9
In a right triangle, c = square root of a² + b². Solve for a.
Show Answer

c = square root of a² + b²

Square both sides.

c² = a² + b²

Subtract b² from both sides.

c² − b² = a²

a is a length, so use the positive square root.

a = square root of c² − b²

Answer: a = square root of c² − b²

10
Solve A = (b + 6)² for b. The quantity b + 6 is positive.
Show Answer

A = (b + 6)²

The quantity b + 6 is squared.

Take the positive square root of both sides.

square root of A = b + 6

Subtract 6 from both sides.

square root of A − 6 = b

b = square root of A − 6

Answer: b = square root of A − 6

11
Solve y = (x − 4)² for x.
Show Answer

y = (x − 4)²

The quantity x − 4 is squared.

Take the square root of both sides. It can be positive or negative.

square root of y = x − 4

or −square root of y = x − 4

Add 4 to both sides.

x = 4 + square root of y

or x = 4 − square root of y

Answer: x = 4 + square root of y or x = 4 − square root of y

12
Solve P = 2(s + 5)² for s. The quantity s + 5 is positive.
Show Answer

P = 2(s + 5)²

The quantity s + 5 is squared.

Divide both sides by 2.

P2 = (s + 5)²

Take the positive square root of both sides.

square root of P2 = s + 5

Subtract 5 from both sides.

s = square root of P2 − 5

Answer: s = square root of P2 − 5

Need help with literal equations?

Book a Session with Amy