Columbia University University

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## Final Answer

attached are the answers.u should give me tip becuase this was very exhaustive work. i could have done it easily with pen and paper.

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

c) Addition/Elimination Method

Multiplying the first equation by 3 on both sides, we get

3𝑥 + 3𝑦 = 15

5𝑥 − 3𝑦 = −23

Adding these equations, we get

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

Using Quadratic formula 𝑥 =

−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

Adding these two equations,

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...