The solution is [tex](-1,5)[/tex] and [tex](7,-3)[/tex]
Step-by-step explanation:
The expression is [tex]x^{2} +y^{2} -x+3y-42=0[/tex] and [tex]x+y=4[/tex]
Using substitution method we can solve the expression.
Let us substitute [tex]x=4-y[/tex] in [tex]x^{2} +y^{2} -x+3y-42=0[/tex]
[tex](4-y)^{2} +y^{2} -(4-y)+3y-42=0[/tex]
Expanding and simplifying the expression, we get,
[tex]\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}[/tex]
Let us use the quadratic equation formula to solve this equation,
[tex]\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}[/tex]
Thus, [tex]y=5[/tex] and [tex]y=-3[/tex]
Substituting y-values in the equation [tex]x+y=4[/tex], we get the value of x.
For [tex]y=5[/tex] ⇒ [tex]x=4-5=-1[/tex]
For [tex]y=-3[/tex] ⇒ [tex]x=4+3=7[/tex]
Thus, the solution set is [tex](-1,5)[/tex] and [tex](7,-3)[/tex]
