Answer:
x = 8, y = 0
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}y=\dfrac{1}{2}x-4&\\y=-\dfrac{1}{4}x+2\end{cases}[/tex]
First, substitute the first equation into the second:
[tex]\implies \dfrac{1}{2}x-4=-\dfrac{1}{4}x+2[/tex]
SOLVE FOR X.
Step 1: Multiply both sides by [tex]4[/tex] and simplify.
[tex]\\\implies 4\left(\dfrac{1}{2}x\right)-4(4)&=4\left(-\dfrac{1}{4}x\right)+2(4)\\\\\implies \dfrac{4}{2}x-16=-\dfrac{4}{4}x+8\\\\\implies 2x-16=-x+8[/tex]
Step 2: Add (1)[tex]x[/tex] to both sides.
[tex]\\\implies 2x + x-16=-x+x+8\\\\\implies 3x-16=8[/tex]
Step 3: Add [tex]16[/tex] to both sides.
[tex]\\\implies3x-16+16=8+16\\\\\implies 3x=24[/tex]
Step 4: Divide both sides by [tex]3[/tex].
[tex]\\\implies \dfrac{3x}{3}=\dfrac{24}{3}\\\\\implies \boxed{x=8}[/tex]
SOLVE FOR Y.
Step 1: Substitute [tex]8[/tex] as the value of [tex]x[/tex] in any of the given equations.
[tex]\\\implies y=\dfrac{1}{2}(8)-4[/tex]
Step 2: Simplify.
[tex]\\\implies y=4-4\\\\\implies \boxed{y=0}[/tex]
The solution to this system of equations is (8,0).
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