Solve the equation u-5 =15 and state the axiom that you use here. -Maths 9th

Solve the equation u-5 =15 and state the axiom that you use here. -Maths 9th
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x = 5 and y = 2 is a solution of the linear equation. -Maths 9th

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Last Answer : (c) (a) Take x + 2y, on putting x=5 and y = 2, we get 5 + 2(2) = 5+ 4 = 9≠7. So, (5, 2) is not a solution of x + 2y = 7 (b) Take 5x + 2y, on putting x = 5 and y = 2, we get 5x 5 + 2 x2 =25+ 4 = ... on putting x = 5 and y = 2, we get 5 5 + 2=25 + 2 =27 ≠7 So, (5, 2) is not a solution of 5x + y = 7.

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Last Answer : From x + y = 5, If x = 0 0 + y = 5 y = 5 Therefore (0,5) If x = 1 1 + y = 5 y =5 - 1 y = 4 Therefore (1,4) Draw a graph for this And From 3x - 2y = 0 If x = 0 3 (0) - 2y = 0 0 - ... 2y = 0 -2y = -6 y = -6/-2 y = 3 Therefore (2,3) Draw a graph for these points And the point of intersection is (2,3)

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Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.