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 Number of inequalities to solve: 23456789
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# Solving Equations &amp; Inequalities

## Solutions to Exercises

Part (d) on the next page.

 given subtract 4/5 both sides arithmetic multiply both sides by 3/2

7.2. Solutions:
(a) Solve for x :

 given add 2 to both sides add −7x to both sides divide by −4

(b) Solve for x:

 given add −3 to both sides add −4x to both sides divide both sides by −1

(c) Solve for x:

 given multiply both sides by 6 expand add −8x to both sides divide both sides by 5

7.3. Answers: Make sure you understand the method of solution .
The answers are given only here.
(a) Solve for x in 5x − 3y = 4.

(b) Solve for y in 5x − 3y = 4.

(c) Solve for z in x2z − 12x + y = 1.

7.4. Solutions:
(a) Solve

 given mulitply both sides by 3x + 8 expand add −x − 40 to both sides transpose divide by 14 reduce fractions

(b) Solve

 given multiply both sides by 3 − 8x expand r .h.s. add 16x − 2 to both sides combine similar terms divide by 21

(c) Solve (2x − 3)2= (2x − 7)2.

 given expand using (5), Lesson 5 add −4x2 to both sides add 28x − 9 to both sides combine divide by 16

Comment: This problem is similar to (a) of Example 7.2; however, in
my solution I gave a more “traditional’ solution. In the second line
above, I simply expanded the binomials —this is perhaps what you
did yourself. The rest follows using standard methods.

(d) Solve

 given multiply both sides by 2x+1 expand using (2) of Lesson 5 add −8x2 to both sides add −6x − 1 to both sides divide both sides by −4

Now what do you think of that!

7.5. Solutions:
(a) Solve for x: x2 − 7x+12 = 0.
x2 − 7x+12 = 0
(x − 3)(x − 4) = 0

therefore, either

Presentation of Solution:

(b) Solve for x: x2 + 3x = 10.
x2 + 3x = 10
x2 + 3x − 10 = 0
(x + 5)(x − 2) = 0

therefore, either

Presentation of Solution:

(c) Solve for x:

mulitply both sides by x2+ 1

therefore, either

(a) 12x2 − 17x+6 = 0

 given factor it !

From this we can see that the solutions are

Presentation of Solutions:

(b) 20x2 + 3x = 2.

 add −2 to both sides factor it!

Thus,

Presentation of Solutions:

(c)

 multiply both sides by x − 1 factor it– perfect squrare !

Presentation of Solution:

(a) x3 − 2x2 − 3x = 0. Factoring this we obtain
x(x2 − 2x − 3) = 0
x(x − 3)(x + 1) = 0
Presentation of Solutions:

(b) x4 − 16 = 0. Let’s factor—difference of two squares!
(x2 − 4)(x2 + 4) = 0
(x − 2)(x + 2)(x2 + 4) = 0
Presentation of Solutions:
cannot be factored.

(c) x4 − 2x2 − 3 = 0. This is a quadratic equation in the variable
x2: (x2)2 −2(x2)−3 = 0. If you don’t understand what I mean,
temporarily put y = x2; our equation becomes y2 − 2y −3 = 0.

This is clearly a quadratic in y , but y = x2, so it is a quadratic
in x2. Let’s factor it using the factoring techniques.

 factor! again!

My the Zero -Product Prinicple, we then have

Presentation of Solutions:

(d) x4 − 5x2 + 6 = 0. This is again quadratic in x2.

 factor! again!

By the Zero- Product Prinicple , we then have
=>
or,
=>
or,
=>
or,
=>
Presentation of Solutions:

7.8. Solutions:
(a) Solve for x: 8x2 − 2x − 1 = 0.

 given Steps 2 & 3 Step 4 perfect square divide by 8 take square root add 1/8to both sides

Presentation of Solutions:
Comments: Here, mysolution uses a slight variation in the techniques
illustrated in the examples. Rather than having all term
on the left-hand side, I took the constant term to the right -hand
side, then when I completed the square, I added 8/64 to both sides
of the equation.

(b) Solve for x: 3x2 + 5x − 2 = 0.

 given Steps 2 & 3 Step 4 perfect square divide by 8 take square root add 1/8to both sides

Presentation of Solution:

(c) Solve for x: x2 + x − 1 = 0.

 given Steps 2 & 3 Step 4 perfect square take square root add −1/2 to both sides

Presentation of Solution: or,
Presentation of Solution:(Verify!)

7.9. Solutions:
(a) Solve for x: 2x2 + 5x − 12 = 0.

Presentation of Solution:

(b) Solve for x: 3x2 − 7x+1 = 0

Presentation of Solutions:

(c) Solve for x: x2 +1 = 0

Therefore, this equation has no solutions .

(d) Solve for x: x2+x = 3. Begin byputting it into the proper form:
x2 + x − 3 = 0.

Presentation of Solutions:

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