If $A = 5 x - 2 y + 4$ , $B = - 4 x + 6 y - 1$ and $C = 11 y - 10$ , find $A + B - C$ .

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Solution

Step 1: Find the value of $A + B$
Given $A = 5 x - 2 y + 4$ , $B = - 4 x + 6 y - 1$ , $C = 11 y - 10$
$A + B = (5 x - 2 y + 4) + (- 4 x + 6 y - 1)$
$A + B = (5 x - 4 x) + (- 2 y + 6 y) + (4 - 1)$
$∴ A + B = x + 4 y + 3$
Step 2: Find the value of $A + B - C$
$A + B - C = (A + B) - C$
$A + B - C = (x + 4 y + 3) - (11 y - 10)$
$A + B - C = x + (4 y - 11 y) + (3 + 10)$
$∴ A + B - C = x - 7 y + 13$
Hence the value of $A + B - C$ is $x - 7 y + 13$

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In a cyclic quadrilateral ABCD, $∠ A = (2 x + 4)^{\circ}, ∠ B = (y + 3)^{\circ}, ∠ C = (2 y + 10)^{\circ}, ∠ D = (4 x - 5)^{\circ}$ . Find the four angles.

Q. If

A = 4 x - 6 y

B = 5 x + 4 y

C = 3 y - 3 c

then find the value of

A + B - C

ii)

2 A - 3 B + 4 C

Q. If

A = 5 x^{2} + 3 x^{2} - 4 x + 1, B = 3 x^{3} + 5 x^{2} + 3 x - 8

and

C = 6 x^{3} - 4 x^{2} - 7 x + 3

find.

(A + B) = C

Q. If

A = 3 x^{2} - 4 x + 1, B = 5 x^{2} + 3 x - 8

and

C = 4 x^{2} - 7 x + 3

, then find:
(i)

(A + B) - C

(ii)

B + C - A

(iii)

A + B + C

Q. If

A = 3 x^{2} - 4 x + 1, B = 5 x^{2} + 3 x - 8

and,

C = 4 x^{2} - 7 x + 3

, then find:
(i) (A + B) – C
(ii) B + C – A
(iii) A + B + C

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If A=5x-2y+4, B=-4x+6y-1 and C=11y-10, find A+B-C.

Step 1: Find the value of A+BGiven A=5x-2y+4, B=-4x+6y-1, C=11y-10A+B=5x-2y+4+-4x+6y-1A+B=5x-4x+-2y+6y+4-1∴A+B=x+4y+3Step 2: Find the value of A+B-CA+B-C=A+B-CA+B-C=x+4y+3-11y-10A+B-C=x+4y-11y+3+10∴A+B-C=x-7y+13Hence the value of A+B-C is x-7y+13

If $A = 5 x - 2 y + 4$ , $B = - 4 x + 6 y - 1$ and $C = 11 y - 10$ , find $A + B - C$ .