Question

3) x2+ y2 - x + 3y - 42 = 0 X+y=4

Answer 1

The solution is $(-1,5)$ and $(7,-3)$

Step-by-step explanation:

The expression is $x^{2} +y^{2} -x+3y-42=0$ and $x+y=4$

Using substitution method we can solve the expression.

Let us substitute $x=4-y$ in $x^{2} +y^{2} -x+3y-42=0$

$(4-y)^{2} +y^{2} -(4-y)+3y-42=0$

Expanding and simplifying the expression, we get,

$\begin{array}{r}{16-8 y+y^{2}+y^{2}-4+y+3 y-42=0} \\{2 y^{2}-4 y-30=0}\end{array}$

Let us use the quadratic equation formula to solve this equation,

$\begin{aligned}y &=\frac{4 \pm \sqrt{16-4(2)(-30)}}{2(2)} \\&=\frac{4 \pm \sqrt{16+240}}{4} \\&=\frac{4 \pm 16}{4} \\y &=1 \pm 4\end{aligned}$

Thus, $y=5$ and $y=-3$

Substituting y-values in the equation $x+y=4$, we get the value of x.

For $y=5$ ⇒ $x=4-5=-1$

For $y=-3$ ⇒ $x=4+3=7$

Thus, the solution set is $(-1,5)$ and $(7,-3)$