Subtract b from each side

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=. c. ax + b. =. c – b. ax. =. c – b. Assume a 0 . Divide each side by a. x. =. a. EXAMPLE 1. Solve a literal equation. Solve ax +b = c for x . Then use the solution to solve 2 x + 5 = 11. SOLUTION. Solve ax + b = c for x. STEP 1. Write original equation. - PowerPoint PPT Presentation

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Subtract b from each side.

Write original equation.

Solve ax + b = c for x.STEP 1

SOLUTION

Solve ax +b = c for x. Then use the solution to solve 2x + 5 = 11.

Solve a literal equation

EXAMPLE 1

xc – b

a=

ax + b = c

ax = c – b

Assume a 0. Divide each side by a.=

The solution of 2x + 5 = 11 is 3.ANSWER

Simplify.

Substitute 2 for a, 5 for b, and 11 for c.

Solution of literal equation.

Use the solution to solve 2x + 5 = 11.STEP 2

Solve a literal equation

EXAMPLE 1

11 – 52=

x = c – b

a

= 3

GUIDED PRACTICE for Example 1

Subtract a from each side.

Write original equation.

Solve a – bx = c for x.STEP 1

SOLUTION

1. Solve a – bx = c for x.

xa – c

b=

a – bx = c

– bx = c – a

Assume b 0. Divide each side by – 1.=

Solve the literal equation for x . Then use the solution to solve the specific equation

GUIDED PRACTICE for Example 1

The solution of 12 – 5x = –3 is 3.ANSWER

Simplify.

Substitute a for 12, –3 for c, and 5 for b.

Solution of literal equation.

Use the solution to solve 12 – 5x = –3.STEP 2

12 – (–3)5=

x = a – c

b

= 3

GUIDED PRACTICE for Example 1

Subtract bx from each side.

Write original equation.

Solve a x = bx + c for x.STEP 1

SOLUTION

2. Solve a x = bx + c for x.

cx

a – b=

a x = bx + c

a x – bx = c

Assume a 0. Divide each=side by a – b.

GUIDED PRACTICE for Example 1

The solution of 11x = 6x + 20. is 4.ANSWER

Simplify.

Solution of literal equation.

Use the solution to solve 11x = 6x + 20.STEP 2

2011 – 6=

x = c

a – b

= 4

Substitute a for 11, 20 for c, and6 for b.

Divide each side by 2.

Write original equation.

Write 3x + 2y = 8 so that y is a function of x.

EXAMPLE 2 Rewrite an equation

Subtract 3x from each side.

3x + 2y = 8

2y = 8 – 3x

32

y = 4 – x

Multiply each side by 2.

Write original formula.

SOLUTION

Use the rewritten formula to find the height of the triangle shown, which has an area of 64.4 square meters.

b.

Solve the formula for the height h.a.

EXAMPLE 3 Solve and use a geometric formula

The area A of a triangle is given by the formula A = bh where b is the base and h is the height.

12

a. bh12A =

2A bh=

Substitute 64.4 for A and 14 for b.

Write rewritten formula.

Substitute 64.4 for A and 14 for b in the rewritten formula.

b.

Divide each side by b.

EXAMPLE 3 Solve and use a geometric formula

2A b

h=

= 2(64.4) 14

= 9.2 Simplify.

ANSWER The height of the triangle is 9.2 meters.

h2A b=

GUIDED PRACTICE for Examples 2 and 3

Divide each side by 4.

Write original equation.

3 . Write 5x + 4y = 20 so that y is a function of x.

Subtract 5x from each side.

5x + 4y = 20

4y = 20 – 5x

54

y = 5 – x

GUIDED PRACTICE for Examples 2 and 3

Divide each side by 2.

Write original equation.

Subtract 2l from each side.

a . p = 2l + 2w

The perimeter P of a rectangle is given by the formula P = 2l + 2w where l is the length and w is the width.

a. Solve the formula for the width w.

4 .

p – 2l = 2w

p – 2l2

= w

SOLUTION

GUIDED PRACTICE for Examples 2 and 3

Simplify.

Write original equation.

Substitute 19.2 for P and 7.2 for l.

Substitute 19.2 for P and 7.2 for l in the rewritten formula

b .

w = p –2l 219.2 – 2 (7.2)

2=

= 2.4

The width of the rectangle is 2.4 feet

EXAMPLE 4 Solve a multi-step problem

You are visiting Toronto, Canada, over the weekend. A website gives the forecast shown. Find the low temperatures for Saturday and Sunday in degrees Fahrenheit. Use the formula C = (F – 32) where C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit.

59

Temperature

Simplify.

Write original formula.

SOLUTION

EXAMPLE 4 Solve a multi-step problem

Multiply each side by , the reciprocal of .

9

55

9

Add 32 to each side.

STEP 1 Rewrite the formula. In the problem,degrees Celsius are given and degrees Fahrenheit need to be calculated. The calculations will be easier if the formula is written so that F is a function of C.

(F – 32)59C =

F – 32C95

=

95C + 32

=F

. (F – 32)95

59C9

5=

EXAMPLE 4 Solve a multi-step problem

ANSWER

95=The rewritten formula is F C + 32.

The low for Saturday is 57.2°F.

ANSWER

= 25.2 + 32

Saturday (low of 14°C)

Find the low temperatures for Saturday and Sunday in degrees Fahrenheit.

EXAMPLE 4 Solve a multi-step problem

STEP 2

Sunday (low of 10°C)

= (14)+ 3295 = (10)+ 32

95

= 18 + 32

= 57.2 = 50

C + 3295 F = F C + 32

95=

The low for Sunday is 50°F.

ANSWER

GUIDED PRACTICE for Example 4

STEP 1 Rewrite the formula. In the problem,degrees Celsius are given and degrees Fahrenheit need to be calculated. The calculations will be easier if the formula is written so that F is a function of C.

Use the information in Example 4 to find the high temperatures for Saturday and Sunday in degrees Fahrenheit.

5.

GUIDED PRACTICE for Example 4

ANSWER

95=The rewritten formula is F C + 32.

Simplify.

Write original formula.

Add 32 to each side.

(F – 32)59C =

F – 32C95

=

95 C + 32=F

. (F – 32)95

59C9

5= Multiply each side by , the

reciprocal of .

9

55

9

GUIDED PRACTICE for Example 4

The High for Saturday is 71.6°F.

ANSWER

= 39.6 + 32

Saturday (High of 22°C)

Find the high temperatures for Saturday and Sunday in degrees Fahrenheit.

STEP 2

Sunday (High of 16°C)

= (22)+ 3295 = (16)+ 32

95

= 28.8 + 32

= 71.6 = 60.8

C + 3295 F =

F C + 3295=

The High for Sunday is 60.8°F.

ANSWER

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