how to do sum number 14
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3x + 4y = 16 & xy = 4
9x² + 16y² = (3x)² + (4y)² = (3x+4y)² - 2*3x*4y = 16² - 24*4 = 256 - 96 = 160
9x² + 16y² = (3x)² + (4y)² = (3x+4y)² - 2*3x*4y = 16² - 24*4 = 256 - 96 = 160
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Given Equation is 3x + 4y = 16.
On Squaring both sides, we get
(3x + 4y)^2 = (16)^2
9x^2 + 16y^2 + 2 * 3x * 4y = 256
9x^2 + 16y^2 + 24(xy) = 256
9x^2 + 16y^2 + 24 * 4 = 256
9x^2 + 16y^2 + 96 = 256
9x^2 + 16y^2 = 256 - 96
9x^2 + 16y^2 = 160.
Hope this helps!
On Squaring both sides, we get
(3x + 4y)^2 = (16)^2
9x^2 + 16y^2 + 2 * 3x * 4y = 256
9x^2 + 16y^2 + 24(xy) = 256
9x^2 + 16y^2 + 24 * 4 = 256
9x^2 + 16y^2 + 96 = 256
9x^2 + 16y^2 = 256 - 96
9x^2 + 16y^2 = 160.
Hope this helps!
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