Question

y = x2 – 3x - 4y = 2x + 10Which of the following ordered pairs (x,y) is a solution to the system of equations above?

Answer 1

This question is a simultaneous equation having a quadratiic equation above and a linear equation below.

substitute the value of y in equation 2 for y in equation 1

recall y = 2x + 10 ( equation 2 )

2x + 10 = x2 - 3x -4 ( since y = 2x + 10 )

`\begin{gathered} x^2\text{ - 3x -2x -4 -10 = 0 ( collecting all to the left side ) } \\ x^2\text{ - 5x -14 = 0 ( a quadratic equation emerge ) } \\ (x^2\text{ + 2x ) - ( 7x - 14 ) = 0 ( since -5x = +2x -7x ) } \\ x\text{ ( x + 2 ) - 7 ( x + 2 ) = 0 ( factorising ) } \\ \text{therefore ( x + 2 ) ( X - 7 ) = 0 ( factorised ) } \\ \text{Therefore the solution for x, is }x+\text{ 2 = 0 or x - 7 = 0} \\ \text{Hence, x = -2 or x = 7 } \end{gathered}`

Having gotten the two values of x, we see that -2 is in the x position in the options given to us but 7 was ignored. Therefore we shall use x = -2 to find the value of y in equation 2

Recall, y = 2x + 10 ( equation 2 )

thus, y = 2 ( -2 ) + 10 ( since x = -2 )

hence, y = -4 + 10 ( since 2 x -2 = _4 )

so, y = +6

finally, x = -2 and y = 6. the best coordinate is ( -2, 6 )