1. 10-5 Solving for a Variable Warm Up Problem of the Day
Course 3
Warm Up
Problem of the Day
Lesson Presentation

2. Solving for a Variable 10-5 Warm Up Solve. 1. 8x – 9 = 23
Course 3
10-5
Solving for a Variable
Warm Up
Solve.
1. 8x – 9 = 23
2. 9x + 12 = 4x + 37
3. 6x – 8 = 7x + 3
4. x + 3 =
x = 4
x = 5
x = –11 1 2 1 8 23 4
x = – , or –5 3

3. Solving for a Variable 10-5 Problem of the Day
Course 3
10-5
Solving for a Variable
Problem of the Day
The formula A = 4r2 gives the surface area of a geometric figure. Solve the formula for r.
Can you identify what the geometric figure is?
sphere

4. Solving for a Variable Learn to solve an equation for a variable. 10-5
Course 3
10-5
Solving for a Variable
Learn to solve an equation for a variable.

5. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation.

6. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
Additional Example 1A: Solving for a Variable by Addition or Subtraction
Solve for the indicated variable.
A. Solve a – b + 1 = c for a.
a – b + 1 = c
Add b and subtract 1 from both sides.
+ b – b – 1
a = c + b – 1
Isolate a.

7. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
Additional Example 1B: Solving for a Variable by Addition or Subtraction
Solve for the indicated variable.
B. Solve a – b + 1 = c for b.
a – b + 1 = c
Subtract a and 1 from both sides.
– a – – a – 1
–b = c – a – 1
Isolate b.
–1  (–b) = –1  (c – a – 1)
Multiply both sides by –1.
b = –c + a + 1
Isolate b.

8. Solving for a Variable 10-5 Try This: Example 1A
Course 3
10-5
Solving for a Variable
Try This: Example 1A
Solve for the indicated variable.
A. Solve y – b + 3 = c for y.
y – b + 3 = c
Add b and subtract 3 from both sides.
+ b – b – 3
y = c + b – 3
Isolate y.

9. Solving for a Variable 10-5 Try This: Example 1B
Course 3
10-5
Solving for a Variable
Try This: Example 1B
Solve for the indicated variable.
B. Solve p – w + 4 = f for w.
p – w + 4 = f
Subtract p and 4 from both sides.
– p – – p – 4
–w = f – p – 4
Isolate w.
–1  (–w) = –1  (f – p – 4)
Multiply both sides by –1.
w = –f + p + 4
Isolate w.

10. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values.

11. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
Additional Example 2A: Solving for a Variable by Division or Square Roots
Solve for the indicated variable. Assume all values are positive.
A. Solve A = s2 for s.
A = s2
√A = √s2 Take the square root of both sides.
√A = s Isolate s.

12. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
Additional Example 2B: Solving for a Variable by Division or Square Roots
Solve for the indicated variable. Assume all values are positive.
B. Solve V = IR for R.
V = IR
= Divide both sides by I. IR I V I V
= R Isolate R.

13. Solving for a Variable 10-5
Course 3
10-5
Solving for a Variable
Additional Example 2C: Solving for a Variable by Division or Square Roots
C. Solve the formula for the area of a trapezoid for h. Assume all values are positive.
A = h(b1 + b2) 1 2
Write the formula.
2 • A = 2 • h(b1 + b2) 1 2
Multiply both sides by 2.
2A = h(b1 + b2) 2A
(b1 + b2)
h(b1 + b2) =
Divide both sides by (b1 + b2). 2A
(b1 + b2)
= h
Isolate h.

14. Solving for a Variable 10-5 Try This: Example 2A
Course 3
10-5
Solving for a Variable
Try This: Example 2A
Solve for the indicated variable. Assume all values are positive.
A. Solve w = g2 + 4 for g.
w = g2 + 4
– – 4 Subtract 4 from both sides.
w – 4 = g2
Take the square root of both sides.
√w – 4 = √g2
√ w – 4 = g Isolate g.

15. Solving for a Variable 10-5 Try This: Example 2B
Course 3
10-5
Solving for a Variable
Try This: Example 2B
Solve for the indicated variable. Assume all values are positive.
B. Solve A = lw for w.
A = lw
= Divide both sides by l. lw l A l A
= w Isolate w.

16. Solving for a Variable 10-5 Try This: Example 2C
Course 3
10-5
Solving for a Variable
Try This: Example 2C
Solve for the indicated variable. Assume all values are positive.
C. Solve s = 180(n – 2) for n.
s = 180(n – 2)
s (n – 2)
180 =
Divide both sides by 180. s
180
= (n – 2)
+ 2
+ 2
Add 2 to both sides. s
180
+ 2 = n

17. Solving for a Variable 10-5 Remember!
Course 3
10-5
Solving for a Variable
To find solutions (x, y), choose values for x substitute to find y.
Remember!

18. Additional Example 3: Solving for y and Graphing
Course 3
10-5
Solving for a Variable
Additional Example 3: Solving for y and Graphing
Solve for y and graph 3x + 2y = 8.
3x + 2y =
–3x –3x x y
2y = –3x + 8 –2 7 4
–3x + 8 2 2y = 2 1
y =
–3x 2 4 –2

19. Additional Example 3 Continued
Course 3
10-5
Solving for a Variable
Additional Example 3 Continued
3x + 2y = 8

20. Solving for a Variable 10-5 Try This: Example 3
Course 3
10-5
Solving for a Variable
Try This: Example 3
Solve for y and graph 4x + 3y = 12.
4x + 3y = x y
–4x – 4x –3 8
3y = –4x + 12 4
–4x + 12 3 3y = 3 6 –4
y =
–4x 3

21. Try This: Example 3 Continued
Course 3
10-5
Solving for a Variable
Try This: Example 3 Continued y 10 8 6
4x + 3y = 12 4 2 x
–4 – –2 –4 –6

22. Insert Lesson Title Here
Course 3
10-5
Solving for a Variable
Insert Lesson Title Here
Lesson Quiz: Part 1
Solve for the indicated variable.
1. P = R – C for C.
2. P = 2l+ 2w for l.
3. V = Ah for h.
4. R = for S.
C = R - P
= l
P – 2w 2
= h 3V A 1 3
C – S t
C – Rt = S

23. Insert Lesson Title Here
Course 3
10-5
Solving for a Variable
Insert Lesson Title Here
Lesson Quiz: Part 2
5. Solve for y and graph 2x + 7y = 14.
y = – 2x 7