INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
math
hands
1. Solve
Solving
rational equations
2
4
=
3
x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
2
4
=
3
x
3x ·
2
4
= 3x ·
3
x
2x = 12
x=6
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
2. Solve
3
−5
=
1
x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
3
−5
=
1
x
x·
3
−5
=x·
1
x
3x = − 5
5
x= −
3
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
3. Solve
pg. 1
−2
12
=
5
x
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
math
hands
Solving
rational equations
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
−2
12
=
5
x
−2
12
= 5x ·
5
x
5x ·
− 2x = 60
x = − 30
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
4. Solve
9
−1
=
5
x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
−1
9
=
5
x
5x ·
(given)
−1
9
= 5x ·
5
x
(to kill denominators)
− x = 45
(BI, cleans up nicely)
x = − 45
(BI)
the student is advised to check final answer by substituting into original equation.
5. Solve
−2
7
=
15
9x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
pg. 2
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
Solving
rational equations
math
hands
−2
7
=
15
9x
45x ·
(given)
−2
7
= 45x ·
15
9x
− 6x = 35
x= −
(to kill denominators)
(BI, cleans up nicely)
35
6
(BI)
the student is advised to check final answer by substituting into original equation.
6. Solve
5
4
=
−6
4x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
4
5
=
−6
4x
12x ·
5
4
= 12x ·
−6
4x
− 10x = 12
6
x= −
5
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
7. Solve
3
−5
=
7
49x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
pg. 3
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
math
hands
3
−5
=
7
49x
49x ·
3
−5
= 49x ·
7
49x
21x = − 5
5
x= −
21
Solving
rational equations
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
8. Solve
−2
7
=
15
25x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
7
−2
=
15
25x
75x ·
−2
7
= 75x ·
15
25x
− 10x = 21
21
x= −
10
(given)
(to kill denominators)
(BI, cleans up nicely)
(BI)
the student is advised to check final answer by substituting into original equation.
9. Solve
−1
9
=
20
16x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. In this case, the LCM is easy to determine, and we proceed as follows:
pg. 4
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
Solving
rational equations
math
hands
−1
9
=
20
16x
80x ·
−1
9
= 80x ·
20
16x
− 4x = 45
x= −
(given)
(to kill denominators)
(BI, cleans up nicely)
45
4
(BI)
the student is advised to check final answer by substituting into original equation.
10. Solve
−2
2
7
+
=
15
45x
9x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 15, 45x, and 9x which can be
primefactorized into 5 · 3,
5 · (3)2 · (x),
and x · (3)2 respectively, from which we can see the LCM to be
(x)(5)(3)2 = 45x
−2
2
7
+
=
15
45x
9x
45x ·
−2
2
7
+ 45x ·
= 45x ·
15
45x
9x
− 6x + 2 = 35
− 6x = 33
33
x=
−6
(given)
(CLM, to kill denominators)
(BI, cleans up nicely)
(BI)
(BI, CLM)
the student is advised to check final answer by substituting into original equation.
11. Solve
pg. 5
3
3
11
+
=
10 20x
4x
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 03 SEC 01 SOLUTIONS
math
hands
Solving
rational equations
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 10, 20x, and 4x which can be
primefactorized into 5 · 2,
5 · (2)2 · (x),
and x · (2)2 respectively, from which we can see the LCM to be
2
(x)(5)(2) = 20x
3
11
3
+
=
10 20x
4x
20x ·
3
3
11
+ 20x ·
= 20x ·
10
20x
4x
(given)
(CLM, to kill denominators)
6x + 3 = 55
(BI, cleans up nicely)
6x = 52
52
x=
6
(BI)
(BI, CLM)
the student is advised to check final answer by substituting into original equation.
12. Solve
3
1
11
+
=
26 338x
169x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 26, 338x, and 169x which can
be primefactorized into 2 · 13,
2 · (13)2 · (x),
and x · (13)2 respectively, from which we can see the LCM
2
to be (x)(2)(13) = 338x
1
11
3
+
=
26 338x
169x
338x ·
3
1
11
+ 338x ·
= 338x ·
26
338x
169x
(given)
(CLM, to kill denominators)
39x + 1 = 22
(BI, cleans up nicely)
39x = 21
21
x=
39
(BI)
(BI, CLM)
the student is advised to check final answer by substituting into original equation.
pg. 6
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 01 SOLUTIONS
13. Solve
Solving
rational equations
2
7
−2
+
= 2
2
15
45x
9x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 15, 45x2 , and 9x2 which can be
primefactorized into 5 · 3,
5 · (3)2 · (x2 ),
and x2 · (3)2 respectively, from which we can see the LCM to be
2
2
2
(x )(5)(3) = 45x
−2
2
7
+
= 2
2
15
45x
9x
45x2 ·
(given)
−2
2
7
+ 45x2 ·
= 45x2 · 2
15
45x2
9x
− 6x2 + 2 = 35
(BI, cleans up nicely)
2
− 6x = 33
33
x2 =
−6
s
x=±
(CLM, to kill denominators)
(BI)
(BI, CLM)
33
−6
(SRP)
the student is advised to check final answer by substituting into original equation.
14. Solve
−2
2
7
+
= 2
15
45x2
9x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 15, 45x2 , and 9x2 which can be
primefactorized into 5 · 3,
5 · (3)2 · (x2 ),
and x2 · (3)2 respectively, from which we can see the LCM to be
(x2 )(5)(3)2 = 45x2
pg. 7
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 01 SOLUTIONS
−2
2
7
+
= 2
15
45x2
9x
45x2 ·
(given)
−2
2
7
+ 45x2 ·
= 45x2 · 2
2
15
45x
9x
− 6x2 + 2 = 35
(CLM, to kill denominators)
(BI, cleans up nicely)
2
− 6x = 33
33
x2 =
−6
s
x=±
Solving
rational equations
(BI)
(BI, CLM)
33
−6
(SRP)
the student is advised to check final answer by substituting into original equation.
15. Solve
5
3
7
+
= 2
2
6 12x
4x
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 6, 12x2 , and 4x2 which can be
primefactorized into 3 · 2,
3 · (2)2 · (x2 ),
and x2 · (2)2 respectively, from which we can see the LCM to be
2
2
2
(x )(3)(2) = 12x
5
3
7
+
= 2
2
6 12x
4x
12x2 ·
(given)
5
3
7
+ 12x2 ·
= 12x2 · 2
2
6
12x
4x
10x2 + 3 = 21
(BI, cleans up nicely)
2
10x = 18
18
x2 =
10
s
x=±
(CLM, to kill denominators)
(BI)
(BI, CLM)
18
10
(SRP)
the student is advised to check final answer by substituting into original equation.
pg. 8
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
math
hands
CH 03 SEC 01 SOLUTIONS
16. Solve
Solving
rational equations
7
4
5
+
=
2
39 507x
169x2
Solution: Usually, an effective idea is to get rid of denominators by multiplying both sides of the equation by
the LCM of denominators. One way to determine the LCM is to primefactorize the denominators so as to see
the prime pieces, from the prime pieces we simply use the definition of the LCM as the product of all primes
each raised to the higher appearing exponent. In this case the denominators are 39, 507x2 , and 169x2 which can
be primefactorized into 3 · 13,
3 · (13)2 · (x2 ),
and x2 · (13)2 respectively, from which we can see the LCM
2
2
2
to be (x )(3)(13) = 507x
7
4
5
+
=
2
39 507x
169x2
507x2 ·
7
4
5
+ 507x2 ·
= 507x2 ·
39
507x2
169x2
91x2 + 4 = 15
x=±
(CLM, to kill denominators)
(BI, cleans up nicely)
2
91x = 11
11
x2 =
91
s
(given)
(BI)
(BI, CLM)
11
91
(SRP)
the student is advised to check final answer by substituting into original equation.
pg. 9
c
2007-2011
MathHands.com v.1010