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Solving Equations & Inequalities
Solutions to Exercises
7.1. Answers:
Part (d) on the next page.
Solution: The most difficult thing about this one is the arithmetic !
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 
Presentation of Answer:
(b) Solve for x:
given  
add −3 to both sides  
add −4x to both sides  
divide both sides by −1 
Presentation of Answer:
(c) Solve for x:
given  
multiply both sides by 6  
expand  
add −8x to both sides  
divide both sides by 5 
Presentation of Answer:
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 x^{2}z − 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 
Presentation of Answer:
(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 
Presentation of Answer:
(c) Solve (2x − 3)^{2}= (2x − 7)^{2}.
given  
expand using (5), Lesson 5  
add −4x^{2} to both sides  
add 28x − 9 to both sides  
combine  
divide by 16  
Presentation of Answer:
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 −8x^{2} to both sides  
add −6x − 1 to both sides  
divide both sides by −4 
Presentation of Answer:
Now what do you think of that!
7.5. Solutions:
(a) Solve for x: x^{2} − 7x+12 = 0.
x^{2} − 7x+12 = 0
(x − 3)(x − 4) = 0
therefore, either
Presentation of Solution:
(b) Solve for x: x^{2} + 3x = 10.
x^{2} + 3x = 10
x^{2} + 3x − 10 = 0
(x + 5)(x − 2) = 0
therefore, either
Presentation of Solution:
(c) Solve for x:
mulitply both sides by x^{2}+ 1
therefore, either
Presentation of Answer:
7.6. Answers:
(a) 12x^{2} − 17x+6 = 0
given  
factor it ! 
From this we can see that the solutions are
Presentation of Solutions:
(b) 20x^{2} + 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:
7.7. Answers:
(a) x^{3} − 2x^{2} − 3x = 0. Factoring this we obtain
x(x^{2} − 2x − 3) = 0
x(x − 3)(x + 1) = 0
Presentation of Solutions:
(b) x^{4} − 16 = 0. Let’s factor—difference of two squares!
(x^{2} − 4)(x^{2} + 4) = 0
(x − 2)(x + 2)(x^{2} + 4) = 0
Presentation of Solutions:
Comments: The last factor x^{2}+4 is an irreducible quadratic—it
cannot be factored.
(c) x^{4} − 2x^{2} − 3 = 0. This is a quadratic equation in the variable
x^{2}: (x^{2})^{2} −2(x^{2})−3 = 0. If you don’t understand what I mean,
temporarily put y = x^{2}; our equation becomes y^{2} − 2y −3 = 0.
This is clearly a quadratic in y , but y = x^{2}, so it is a quadratic
in x^{2}. Let’s factor it using the factoring techniques.
factor!  
again! 
My the Zero Product Prinicple, we then have
Presentation of Solutions:
(d) x^{4} − 5x^{2} + 6 = 0. This is again quadratic in x^{2}.
factor!  
again! 
By the Zero Product Prinicple , we then have
=>
or,
=>
or,
=>
or,
=>
Presentation of Solutions:
7.8. Solutions:
(a) Solve for x: 8x^{2} − 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 lefthand 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: 3x^{2} + 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: x^{2} + 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: 2x^{2} + 5x − 12 = 0.
Presentation of Solution:
(b) Solve for x: 3x^{2} − 7x+1 = 0
Presentation of Solutions:
(c) Solve for x: x^{2} +1 = 0
Therefore, this equation has no solutions .
(d) Solve for x: x^{2}+x = 3. Begin byputting it into the proper form:
x^{2} + x − 3 = 0.
Presentation of Solutions:
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