Question

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

Answer 1

Calculate the value of $x$ and $y$

Given: $(x^2-3x,y^2+4y)\&(-2,5)$ are equal.

So,

$$\begin{gathered} x^2-3x=-2------\left(1\right) \\ and \\ {y}^2+4y=5------\left(2\right) \end{gathered}$$

Step $1$:-

Equation $\left(1\right)$ becomes ,

$x^2-3x+2=0$

By splitting the middle term in equation $\left(1\right)$ factorization we get,

$$\begin{gathered} x^2-x-2x+2=0 \\ \Rightarrow x(x-1)-2(x-1)=0 \\ \Rightarrow (x-2)(x-1)=0 \\ \Rightarrow x=2orx=1 \end{gathered}$$

Step $2$:-

Equation $\left(2\right)$ becomes ,

$y^2+4y-5=0$

By splitting the middle term in equation $\left(2\right)$ factorization we get,

$$\begin{gathered} y^2+4y-5=0 \\ \Rightarrow y^2–y+5y-5=0 \\ \Rightarrow y(y-1)+5(y-1)=0 \\ \Rightarrow (y-1)(y+5)=0 \\ \Rightarrow y=1ory=-5 \end{gathered}$$

Hence, If the ordered pairs $(x^2-3x,y^2+4y)\&(-2,5)$ are equal, then $x=2orx=1andy=1ory=-5.$