Question

Solve x+y+10 = 0
and
x-Y
+3=0
​

Answer 1

Given equations

$x+y+10=0$

$\implies x+y=-10--(1)$

$x-y+3=0$

$\implies x-y=-3--(2)$

Solving (1) and (2), we get,

$x+y=-10\\\underline{x-y=-3}\\\underline{\underline{2x=-13}}$

$\implies x = \dfrac{-13}{2}$

Substituting the value of x in the second equation, we get,

$\implies \dfrac{-13}{2} -y=-3$

$\implies -y = -3 + \dfrac{13}{2}$

$\implies -y = \dfrac{-6+13}{2}$

$\implies -y = \dfrac{7}{2}$

$\implies y = \dfrac{-7}{2}$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ x \ and \ y \ are \ \frac{-13}{2} \ and \ \frac{-7}{2} \ respectively}}}}}}$

                                                       

Answer 2

Answer:

$\sf{The \ value \ of \ x \ is \ \frac{-13}{2} \ and \ y \ is \ \frac{-7}{2}.}$

Given:

• The given equations are

1. x+y+10=0
2. x-y+3=0

To find:

• The value of x and y.

Solution:

$\sf{The \ given \ equations \ are}$

$\sf{x+y=-10...(1)}$

$\sf{x-y+3=0}$

$\sf{\therefore{x-y=-3...(2)}}$

$\sf{Add \ equations \ (1) \ and \ (2), \ we \ get}$

$\sf{x+y=-10}$

$\sf{+}$

$\sf{x-y=-3}$

______________

$\sf{2x=-13}$

$\boxed{\sf{\therefore{x=\frac{-13}{2}}}}$

$\sf{Substitute \ x=\frac{-13}{2} \ in \ equation (1)}$

$\sf{\frac{-13}{2}+y=-10}$

$\sf{\therefore{y=-10+\frac{13}{2}}}$

$\sf{\therefore{y=\frac{-20+13}{2}}}$

$\boxed{\sf{\therefore{y=\frac{-7}{2}}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ \frac{-13}{2} \ and \ y \ is \ \frac{-7}{2}.}}}$