Question

If the ordered pairs (x^2-3x , y^2+4y) and (-2,5) are equal ,then find x and y

Answer 1

${x}^{2} - 3x = - 2 \\ {x {}^{2} - 3x }^{} + 2 = 0$

$(x - 1)(x - 2) = 0$

$x = 1 \: \: or \: \: 2$

And

$y {}^{2} + 4y = 5$

${y}^{2} + 4y - 5 = 0 \\ (y - 1)(y + 5) = 0 \\ y = 1 \: \: \: or \: \: \: - 5$

Answer 2

x = 1 or x = 2 and y = 1 or y = -5

Solution:

Given that,

Ordered pairs (x^2-3x , y^2+4y) and (-2,5) are equal

Which means,

$x^2 - 3x = -2 \\\\y^2 + 4y = 5$

Solve them

Case 1:

$x^2 - 3x + 2 = 0 \\\\Split\ the\ middle\ term \\\\x^2 - x - 2x + 2 = 0\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(x^2-x\right)+\left(-2x+2\right) = 0\\\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-x\mathrm{:\quad } \\\\x(x - 1) + (-2x + 2) = 0\\\\\mathrm{Factor\:out\:}-2\mathrm{\:from\:}-2x+2\\\\x\left(x-1\right)-2\left(x-1\right) = 0\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\\\\left(x-1\right)\left(x-2\right) = 0\\\\Thus\\\\x = 1\ or\ x = 2$

Case 2:

$y^2 + 4y = 5 \\\\y^2 + 4y - 5 = 0\\\\Split\ the\ middle\ term \\\\y^2 - y + 5y - 5 = 0\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(y^2-y\right)+\left(5y-5\right) = 0\\\\\mathrm{Factor\:out\:}y\mathrm{\:from\:}y^2-y\mathrm{:\quad } \\\\y(y-1) + (5y-5) = 0 \\\\\mathrm{Factor\:out\:}5\mathrm{\:from\:}5y-5\mathrm{:\quad }\\\\y\left(y-1\right)+5\left(y-1\right) = 0\\\\\mathrm{Factor\:out\:common\:term\:}y-1\\\\\left(y-1\right)\left(y+5\right) = 0\\\\Thus\\\\y = 1\ or\ y = -5$

Result : x = 1 or x = 2 and y = 1 or y = -5

Learn more:

If P + Q = 1 and the ordered pair (p, q) satisfies 3x ➕ 2y = 1 then it also satisfies a) 3x ➕ 4y = 5 b) 5x ➕ 4y = 4 c) 5x ➕ 5y = 4

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Find the solution of the pair of equations x/10+ y/5 - 1 = 0 and x/8 + y/6 = 15 . Hence , find λ , if y = λx + 5.

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