Question

1. solve it: (10x2+3y2+4xy)-(3x2+5y2-3xy)

2. what should be added to
-3a+7b-16 to get 4a-3b+19

3. Find the value of below algebraic expression for x=2,y=1 and z=2

i- 4x2-3y2 + 5z2

ii- 3x2-2x + 4yz+5x2
​

Answer 1

Answer:

Ans 1 10x2 + 3y2 +4xy -3x2 -5y2 +3xy

10x2 -3x2 +3y2 -5y2 +4xy +3xy

7x2 -2y2 +7xy

Ans 2 (4a -3b +19) - (-3a +7b -16)

4a -3b +19 +3a -7b +16

4a +3a -3b -7b +19 +16

7a -10b +25

Answer 2

Answer:

1. Solve: (10x²+3y²+4xy)-(3x²+5y²-3xy)

= 10x²+3y²+4xy-3x²-5y²+3xy {bracket open}

= 10x²-3x²+3y²-5y²+4xy+3xy {re-aranged}

= 7x²-2y²+7xy

2. whatever comes after subtracting -3a+7b-16 from 4a-3b+19 will be the answer.

so, (4a-3b+19) – (-3a+7b-16)

= 4a-3b+19+3a-7b+16. {bracket open}

= 4a+3a-3b-7b+19+16. {rearranged}

= 7a-10b+35

3. Find Value

Given:

x=2

y=1

z=2

(i) 4x²-3y²+5z²

4(2)²–3(1)²+5(2)²

= 4(4)–3(1)+5(4)

= 16–3+20

= 33

(ii) 3x²–2x+4yz+5x²

3x²+5x²–2x+4yz. {rearranged}

= 8x²–2x+4yz

= 8(2)²–2(2)+4(1)(2)

= 8(4)–4+4(2)

= 32–4+8

= 36