Solving Equations
Objectives:
1) To solve equations
2) To solve problems by writing
equations
Solution to An Equation
A number that makes the equation true is a solution
to the equation.
Ex: 5x + 23 = 88
-23
-23
5x
= 65
5
5
x
= 13 Now, check your answer
5(13) + 23 = 88
Since x = 13 satisfies as an answer
to 5x + 23 = 88, we call 13 the
65 + 23 = 88
solution
88
= 88
Example #2: Using the Distributive
Property
Solve 3x – 7(2x – 13) = 3(-2x + 9)
3x – 14x + 91= -6x + 27
-11x + 91 = -6x + 27
+11x
+11x
91 = 5x + 27
- 27
- 27
64 = 5x
5
5
12.8 = x
Example #3: Solving a Formula for
One of its Variables
The formula for the area of a trapezoid is
A = ½ h(b1+b2). Solve the formula for h
2(A) = 2( ½) h(b1+b2)
2A = h (b1+b2)
(b1+b2) (b1+b2)
2A/ (b1+b2) = h
Example #4: Solving an Equation for
One of its Variables
Solve x/a + 1 = x/b for any restrictions on a & b
ab(x/a) + (ab)(1) = (ab)(x/b)
bx + ab = ax
ab = ax – bx
ab = (a – b)x
(a-b) (a-b)
ab/(a-b) = x
Example #5 Word Problems
A dog kennel owner has 100 ft of fencing to
enclose a rectangular dog run. She wants it to
be 5 times as long as it is wide. Find the
dimensions of the dog run.
1st , remember the formula for the perimeter of a
rectangle. P = 2w + 2l or P = 2(w + l)
Since the length is 5 times the width, we’ll
represent everything in terms of width.
w = width
5w = length
Example #5 Word Problems
Write the equation:
2w + 2(5w) = 100
2w + 10w = 100
12w = 100
12
12
w
= 8 1/3
Remember that w represents the width, so
5w = 41 2/3 is the length
Example #6: Using Ratios
The lengths of the sides of a triangle are in the
ratio 3:4:5. The perimeter of the triangle is
18in. Find the lengths of the sides.
Use the ratio and the perimeter to create an
equation:
3x + 4x + 5x = 18
12x = 18
12
12
Example #6: Using Ratios
x = 1.5
Now that you know what x is, it’s time to find
3x, 4x, 5x
3x = 3(1.5) = 4.5
4x = 4(1.5) = 6
5x = 5(1.5) = 7.5
Example #7: Word Problems
Radar detected an unidentified plane 5000mi
away, approaching at 700mi/h. Fifteen minutes
later an interceptor plane was dispatched,
traveling at 800mi/h. How long did the
interceptor take to reach the approaching
plane?
Example #7: Word Problems
Find a relationship in the information.
Distance for approaching plane + distance for the
interceptor = 5000mi
Let t = the time in hours for the interceptor
t + 0.25 = the time in hours for the approaching
plane.
800 t + 700(t + 0.25) = 5000
Example #7: Word Problems
800 t + 700 (t + 0.25) = 5000
800t +700t + 175 = 5000
1500t + 175 = 5000
-175 -175
1500t = 4825
1500 1500
t = 3.217 or 3hr 13min