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Directions:1-5) **In the following questions, two equations I and II are given. Solve both the equations** **and** g**ive Answer**

1.

I.(25 / p^{2}) – (12 / p) + (9 / p^{2}) = (4 / p^{2})

II.9.84 – 2.64 = 0.95 + q^{2}

2.

I. √(900)p + √(1296) = 0

II.(256)^{1/4}q + (216)^{1/3} = 0

3.

I.[(3)^{5} + (7)^{3}]/ 3 = p^{3 }II.7q^{3} = – (15 × 2) + 17q^{3} ^{ }

4.

I.(p^{1/4} / 16)^{2} = 144 / p^{3/2 }II.q^{1/3 } × q^{2/3} × 3104 = 16 × q^{2}

5.

I.3p^{2} – 19p +28 = 0

II.5q^{2} – 18q + 16 = 0

Directions:6-10) **In each question, two equations are given. Find p and q and give the answer:**

6.

a)(p³ – 12p^{2} – p + 12)/(p+1) = 0

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

7.

a)q = 2p + 1

b)2q = 3p – 1

8.

a)9p² – 29p + 22 = 0

b)q² – 7q + 12 = 0

9.

a)3p² – 4p – 32 = 0

b)12q² – 109q + 247 = 0

10.

a)4p + 7q = 42

b)3p – 11q = -1

**Check your Answers below:**

Directions:1-5) **In the following questions, two equations I and II are given. Solve both the equations** **and** g**ive Answer**

##### 1. Question

I.(25 / p^{2}) – (12 / p) + (9 / p^{2}) = (4 / p^{2})

II.9.84 – 2.64 = 0.95 + q^{2}

I.(25/p

^{2}) + (9/p

^{2 }) – (4/p

^{2})

^{ }= (12/p)

(25 + 9 – 4) / p^{2} = 12/p = 30/p^{2} = 12/p

12p = 30

p = 30 / 12 = 5/2 = 2.5

II.9.84 -2.64 = 0.95 + q^{2}

7.2 – 0.95 = q^{2}

q = √(6.25) = ± (2.5)

Clearly p ≥ q

##### 2. Question

- √(900)p + √(1296) = 0
- (256)
^{1/4}q + (216)^{1/3}= 0

I.√(900)p + √(1296) =0

√(900)p = -√(1296)

30p = -36

p = -36 / 30 = -1.2

II.(256)^{1/4} q = (216)^{1/3}

(4^{4})^{1/4} q = – (6^{3})^{1/3} à 4q = -6

Q = -(6/4) = -1.5

Clearly, p > q

##### 3. Question

I.[(3)^{5} + (7)^{3}]/ 3 = p^{3 }II.7q^{3} = – (15 × 2) + 17q^{3} ^{ }

I.[(3)

^{5}+ (7)

^{3}] / 3 = p

^{3}

(243 + 343) / 3 = p^{3}

(586 / 3) = p^{3} à p = 5.8

II.7q^{3} = -30 + 17q^{3} = 10q^{3} = 30

q^{3 }= 30/10 = 3 à q = 1.44

Clearly, p > q

##### 4. Question

- (p
^{1/4}/ 16)^{2}= 144 / p^{3/2} - q
^{1/3 }× q^{2/3}× 3104 = 16 × q^{2}

I.(p

^{1/4}/ 16)

^{2}= (144 / p

^{3/2}) = (p

^{1/2}/ 256) = (144 / p

^{3/2})

(p^{1/2} ) × (p^{3/2} ) = 256 × 144

p^{2} = (256 × 144)

p = √(256 × 144)

p = ± (16 × 12) = ±192

II.q^{1/3} × q^{2/3} × 3104 = 16q^{2}

q × 3104 = 16q^{2}

3104 = 16q

Q = 3104 / 16 = 194

Clearly, p < q

##### 5. Question

I.3p^{2} – 19p +28 = 0

II.5q^{2} – 18q + 16 = 0

I.3p

^{2}– 9p + 28 = 0

3p^{2} – 12p – 7p + 28 = 0

3p (p – 4) – 7 (p – 4) = 0

(p – 4) (3p – 7) = 0

p = 4, 7/3

II.5q^{2} – 18q + 16 = 0

5q^{2} – 10q – 8q + 16 = 0

5q (q – 2) – 8 (q – 2) = 0

(q – 2) (5q – 8) = 0

Q = 2, 8/5

Clearly, p > q

Directions:6-10) **In each question, two equations are given. Find p and q and give the answer:**

##### 6. Question

a)(p³ – 12p^{2} – p + 12)/(p+1) = 0

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

a)(p³ – 12p

^{2}– p + 12)/(p+1) = 0

[p^{2} (p-12) -1(p-12)]/(p+1) = 0

[(p-1)(p+1)(p-12)]/(p+1) = 0

p² – 13p + 12 = 0

Solving, p = 12, 1

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

(q³ + 7q² – 2q² -14q + 12q – 24)/(q-2) = 0

(q-2)(q² + 7q + 12)/(q-2) = 0

Solving, q = -4, -3

Thus, p>q

##### 7. Question

a)q = 2p + 1

b)2q = 3p – 1

Solving the two using substitution method,

p = -3

q = -5

Thus, p>q

##### 8. Question

a)9p² – 29p + 22 = 0

b)q² – 7q + 12 = 0

a)9p² – 18p – 11p + 22 = 0

9p(p-2) – 11(p-2) = 0

(9p-11)(p-2) = 0

p = 11/9, 2.

b)q² – 4q – 3q + 12 = 0

q(q-4) – 3(q-4) = 0

(q-4)(q-3) = 0

q = 3, 4

Thus, p<q

##### 9. Question

a)3p² – 4p – 32 = 0

b)12q² – 109q + 247 = 0

a)3p² – 12p + 8p – 32 = 0

3p(p-4) + 8(p-4) = 0

(3p+8)(p-4) = 0

p = -8/3, 4

b)12q² – 52q – 57q + 247 = 0

4q (3q – 13) – 19(3q – 13) = 0

(4q-19)(3q-13) = 0

q = 19/4, 13/3

Thus, p<q

##### 10. Question

a)4p + 7q = 42

b)3p – 11q = -1

Sol. Solving the two equations using the substitution method,

p = 7

q = 2

Thus, p>q