 Mathematics
Columbia University Chapter 7 Systems and Inequalities Questions Exercises

Columbia University University

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Section 7.1- 7.3
Problem 1 –
𝑥+𝑦 =5
5𝑥 − 3𝑦 = −23
a) As the graphs are distinct nonparallel lines. Hence there is one solution of the system.
b) Substitution method
𝑦 = 5−𝑥
Substituting y in second equation, we get
5𝑥 − 3(5 − 𝑥) = −23
Solving this equation, we get,
𝑥 = −1
Putting x in the previous equation we get,
𝑦 = 5−𝑥
𝑦=6

Multiplying the first equation by 3 on both sides, we get
3𝑥 + 3𝑦 = 15
5𝑥 − 3𝑦 = −23
8𝑥 = −8

𝑥 = −1
Back substituting the value of x in the equation,
𝑦=6
Problem 2 –
𝑥2 + 𝑦2 = 2
𝑦 =𝑥+2
a) Substituting 𝑦 = 𝑥 + 2 in first equation
𝑥 2 + (𝑥 + 2)2 = 2
Solving (𝑥 + 2)2
𝑥 2 + (𝑥 2 + 22 + (2 × 𝑥 × 2)) = 2
Solving
𝑥 2 + (𝑥 2 + 22 + 4𝑥) = 2
2𝑥 2 + 4𝑥 + 4 = 2
2(𝑥 2 + 2𝑥 + 2) = 2
𝑥 2 + 2𝑥 + 2 = 1
𝑥 2 + 2𝑥 + 1 = 0
(𝑥 + 1)(𝑥 + 1) = 0
𝑥+1=0
𝑥 = −1
𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = −1 𝑖𝑛𝑡𝑜 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑦 = (−1) + 2

𝑦=1
Hence the solution set is (-1,1)
b)

Problem 3 –
𝑥2 + 𝑦2 =

3
2

𝑦√2 = 𝑥 2
a)
b) given,
𝑦√2 = 𝑥 2
Substituting the value of 𝑥 2 into the first equation
𝑦√2 + 𝑦 2 =

3
2

2𝑦√2 + 2𝑦 2 = 3
2𝑦√2 + 2𝑦 2 − 3 = 0
2𝑦 2 + 2𝑦√2 − 3 = 0

−𝑏±√𝑏 2 −4𝑎𝑐
2𝑎

Putting values for 𝑥, 𝑎 𝑎𝑛𝑑 𝑏
−2√2 ± √(2√2)2 − (4 × 2 × −3)
𝑦=

2×2

Solving above equation

𝑦=

−2√2 ± √8 − (−24)
2×2
𝑦=

−2√2 ± 4√2
4

𝑦=

√2(−2 ± 4)
4

𝑦=

√2(−2 ± 4)
4

𝑦=

√2(−2 ± 4)
4

Acc to graph, removing negative,

𝑦=

√2
2

Putting the value of y into second equation
√2
× √2 = 𝑥 2
2
Solving
𝑥 = √1
Hence there are 2 set of solutions
c) (1,

√2
),
2

(-1,

√2
)
2

Problem 4 –
𝑥 = 𝑦2
𝑥 = 4 − 𝑦2
a) –
b) Given
𝑥 = 4 − 𝑦2
Putting the value of 𝑦 2 in equation 1
𝑥 = −(𝑥 − 4)
𝑥 = −𝑥 + 4
Hence
𝑥=2
Putting this value in equation 2
2 = 4 − 𝑦2

𝑦2 = 2
Hence y = (±√2)
So the system have 2 possible solution sets
c) (2, √2), (2, −√2)

Problem 5 –
𝑥2 + 𝑦 = 0
𝑦 − 𝑥2 = 1
a)

b) As these two graphs do not intersect each other hence there is no solution
c) Given

𝑦 = 1 + 𝑥2
Putting it in first equation
𝑥2 + 1 + 𝑥2 = 0
2𝑥 2 = −1
𝑥2 = −

1
2

It has a non-real solution for x hence there is no solution
Problem 6 –
a)
𝑦 = 3𝑥 2 − 𝑥
𝑦 = 2𝑥 2 − 3𝑥 + 8
Putting the value of y from first equation to second
3𝑥 2 − 𝑥 = 2𝑥 2 − 3𝑥 + 8
Solving
3𝑥 2 − 2𝑥 2 + 3𝑥 − 𝑥 − 8 = 0
𝑥 2 + 2𝑥 − 8 = 0
(𝑥 + 4)(𝑥 − 2) = 0
Hence
𝑥 = (−4,2)
Putting the value 𝑥,

𝑦 = 3𝑥 2 − 𝑥

When 𝑥 = −4
𝑦 = 3(−4)2 − (−4)
𝑦 = 52
When 𝑥 = 2
𝑦 = 3(2)2 − (2)
𝑦 = 10
Solution sets ( -4,52), (2,10)
b)
𝑥 2 + 4𝑦 2 = 2
3𝑥 − 2𝑦 = −4
From second equation
𝑦 = −(−4 − 3𝑥)
𝑦=

3𝑥 + 4
2

Putting the value of y in first equation
𝑥 2 + 4(
𝑥 2 + 4(

3𝑥 + 4 2
) =2
2

9𝑥 2 16 24𝑥
+
+
)=2
4
4
4

10𝑥 2 + 24𝑥 + 14 = 0
5𝑥 2 + 12𝑥 + 7 = 0
Factoring above,
7
(𝑥 + ) (𝑥 + 1) = 0
5

7

Hence x = -5 𝑎𝑛𝑑 − 1
7

When x = -5
7
3( − ) + 4
5
𝑦=
2
𝑦=−

1
10

When x= -1
𝑦=

3( −1) + 4
2
𝑦=

1
2

c)
𝑥2 + 𝑦2 = 1
𝑥2 − 𝑦2 = 4
𝑥2 + 𝑦2 + 𝑥2 − 𝑦2 = 5
2𝑥 2 = 5
Hence
𝑥2 =

5
2

Putting this value into equation gives no real value for y. Hence there is no possible
solution

Section 7.4
Problem 1 –

a)

1
(𝑥+4)(𝑥−3)(𝑥 2 +1)

Partial Fraction Decomposition for this will be
𝐴

𝐵

𝐶𝑥+𝐷

+
+ (𝑥 2
(𝑥+4) (𝑥−3)

+1)

A, B, C, D ∈ 𝑹
b)

𝑥+5
𝑥 3 (𝑥−1)2 (𝑥 2 +3)2

Partial Fraction Decomposition for this will be
𝐴
𝑥

+

𝐵
𝑥2

+

𝐶
𝑥3

+

𝐷
(𝑥−1)

+

𝐸
(𝑥−1)2

+

𝐹𝑥+𝐺
𝑥 2 +3

𝐻𝑥+𝐼

+ (𝑥 2

+3)2

A, B, C, D, E, F, G, H, I ∈ 𝑹
c)

3𝑡 2 +2𝑡−2
𝑡 2 (2𝑡+5)3 (2𝑡 2 +5)(𝑡 2 +𝑡+1)

Partial Fraction Decomposition for this will be
𝐴
𝑡

+

𝐵
𝑡2

+

𝐶
2𝑡+5

𝐷

+ (2𝑡+5)2 +

𝐸
(2𝑡+5)3

+

𝐹𝑡+𝐺
2𝑡 2 +5

A, B, C, D, E, F, G, H, I ∈ 𝑹

Question 2 –

a)

3𝑥−5
𝑥 2 −5𝑥+6

Partial Fraction Decomposition for this will be

+

𝐻𝑡+𝐼
𝑡 2 +𝑡+1

𝐴
(𝑥−3)

𝐵

+ (𝑥−2)

Comparing the equations, we get
3𝑥 − 5 = 𝐴(𝑥 − 2) + 𝐵(𝑥 − 3)
𝑇ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑜𝑙𝑑𝑠 𝑔𝑜𝑜𝑑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥. Hence substituting x as 2, we get
3 ∗ 2 − 5 = 𝐴(2 − 2) + 𝐵(2 − 3)
𝐵 = −1
Similarly substituting x as 3, we get
3 ∗ 3 − 5 = 𝐴(3 − 2) + 𝐵(3 − 3)
𝐴=4
Hence, the partial fraction decomposition is,
4

1

(𝑥−3) (𝑥−2)
b)

2𝑥 2 −3𝑥+19
𝑥 3 +4𝑥 2 −7𝑥−10

Partial Fraction Decomposition for this will be
𝐴

𝐵

𝐶

+
+ (𝑥−2)
(𝑥+5) (𝑥+1)
Comparing the equations, we get
2𝑥 2 − 3𝑥 + 19 = 𝐴(𝑥 + 1)(𝑥 − 2) + 𝐵(𝑥 + 5)(𝑥 − 2) + 𝐶(𝑥 + 5)(𝑥 + 1)
𝑇ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ℎ𝑜𝑙𝑑𝑠 𝑔𝑜𝑜𝑑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥. Hence substituting x as 2, we get
2 ∗ 2 ∗ 2 − 3 ∗ 2 + 19 = 𝐶(2 + 5)(2 + 1)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡 𝐶 = 1
Similarly substituting x as -1 we get,
2 ∗ (−1) ∗ (−1) − 3 ∗ (−1) + 19 = 𝐵(−1 + 5)(−1 + 1)

Solving the equation, we get B = -2
Similarly substituting x as -5, we get
2 ∗ (−5) ∗ (−5) − 3 ∗ (−5) + 19 = 𝐴(−5 + 1)(−5 − 2)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑒�... manankul (1815)
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