From the sum of -x2 minus-y2- 1- and -2x2- 3y2 minus- 3- subtract the sum of -x2- 4- and -y2 minus- 5-

Let #160;P=x2 y2+1; #160;Q=2x2+3y2 3; #160;R=x2+4; #160;S=y2 5.P+Q=x2 y2+1+2x2+3y2 3=x2 y2+1+2x2+3y2 3=x2+2x2 y2+3y2+1 3=3x2+2y2 2R+S=x2+4+y2 5=x2+4+y2 ...

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Byju's Answer

Standard IX

Mathematics

Addition of Polynomials

From the sum ...

Question

From the sum of (x²−y²+ 1) and (2x²+ 3y²− 3), subtract the sum of (x²+ 4) and (y²− 5).

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Solution

$Let P = (x^{2} - y^{2} + 1); Q = (2 x^{2} + 3 y^{2} - 3); R = (x^{2} + 4); S = (y^{2} - 5) . P + Q = (x^{2} - y^{2} + 1) + (2 x^{2} + 3 y^{2} - 3) = x^{2} - y^{2} + 1 + 2 x^{2} + 3 y^{2} - 3 = x^{2} + 2 x^{2} - y^{2} + 3 y^{2} + 1 - 3 = 3 x^{2} + 2 y^{2} - 2 R + S = (x^{2} + 4) + (y^{2} - 5) = x^{2} + 4 + y^{2} - 5 = x^{2} + y^{2} + 4 - 5 = x^{2} + y^{2} - 1 Now, (P + Q) - (R + S) = (3 x^{2} + 2 y^{2} - 2) - (x^{2} + y^{2} - 1) = 3 x^{2} + 2 y^{2} - 2 - x^{2} - y^{2} + 1 = 3 x^{2} - x^{2} + 2 y^{2} - y^{2} - 2 + 1 = 2 x^{2} + y^{2} - 1$

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From the sum of (x2−y2+ 1) and (2x2+ 3y2− 3), subtract the sum of (x2+ 4) and (y2− 5).

Let P=x2-y2+1; Q=2x2+3y2-3; R=x2+4; S=y2-5.P+Q=x2-y2+1+2x2+3y2-3=x2-y2+1+2x2+3y2-3=x2+2x2-y2+3y2+1-3=3x2+2y2-2R+S=x2+4+y2-5=x2+4+y2-5=x2+y2+4-5=x2+y2-1Now, P+Q-R+S=3x2+2y2-2-x2+y2-1=3x2+2y2-2-x2-y2+1=3x2-x2+2y2-y2-2+1=2x2+y2-1

From the sum of (x²−y²+ 1) and (2x²+ 3y²− 3), subtract the sum of (x²+ 4) and (y²− 5).