Option D:
The solution to the system of equations is (8, –12).
Solution:
Given system of equations are
[tex]$\frac{1}{4} x-\frac{1}{2} y=8[/tex] – – – – (1)
[tex]$\frac{1}{2} x+\frac{3}{4} y=-5[/tex] – – – – (2)
Lets simplify these equations.
In equation (1), take LCM of the denominator and make the same.
[tex]$\frac{1}{4} x-\frac{2}{4} y=8[/tex]
Do cross multiplication, we get
[tex]x-2y=32[/tex]
Add 2y on both sides of the equations.
[tex]x=32+2y[/tex] – – – – (3)
In equation (1), take LCM of the denominator and make the same.
[tex]$\frac{2}{4} x+\frac{3}{4} y=-5[/tex]
Do cross multiplication, we get
[tex]2x+3y=-20[/tex] – – – – (4)
Substitute equation (3) in (4), we get
[tex]2(32+2y)+3y=-20[/tex]
[tex]64+4y+3y=-20[/tex]
[tex]64+7y=-20[/tex]
Subtract 64 from both sides of the equation.
[tex]7y=-84[/tex]
Divide by –7 on both sides of the equation.
y = –12
Substitute y = –12 in equation (3).
[tex]x=32+2(-12)[/tex]
[tex]x=32-24[/tex]
x = 8
The solution to the system of equations is (8, –12).
Hence option D is the correct answer.
