Answer:
(-36,-192)
Step-by-step explanation:
First we are going to get rid of our fractions so that we don't have to worry about them. For the first equation multiply by 12 and so:
[tex]12 \times (\frac{1}{12}x-\frac{1}{12}y=13)\\\\x-y=156[/tex]
Similarly, we will also get rid of our fraction for the second equation by multiplying all of it by 6 and so:
[tex]6 \times (\frac{1}{3}x-\frac{1}{6}y=20)\\\\2x-y=120[/tex]
Now it's much easier to work with these equations. We will use the elimination method to isolate x and thus solve for x by subtracting equation 1 from equation 2 and so:
[tex]2x-y=120\\-(x-y=156)\\\\x=-36[/tex]
Now since x=-36 we will plug that value into any of the new equations to evaluate the y-coordinate of our intersection, I'll choose the second and so:
[tex]2x-y=120\\\\2(-36)-y=120\\\\-72-y=120\\\\-y=120+72\\\\-y=192\\\\y=-192[/tex]
Therefore, the solution for the system of equations is when x = -36 and y = -192. In an ordered pair (x,y) the solution is (-36,-192).
The solution to the system of equation is expressed as (-36, -192)
Given the systems of equations as shown:
[tex]\frac{1}{12} x-\frac{1}{12}y=13\\\frac{1}{3}x-\frac{1}{6}y= 20[/tex]
Using the elimination method as shown:
Multiplying equation 1 by 12 and equation 2 by 3 to have:
[tex]\frac{12}{12} x-\frac{12}{12}y=156\\\frac{3}{3}x-\frac{3}{6}y= 60[/tex]
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Subtract both equations:
-y - (-0.5y) = 156 - 60
-y+0.5y = 96
-0.5y = 96
y = 96/-0.5
y = -192
Substitute y = -192 into equation 1 to get x;
x/12 + 192/12 = 13
x/12 + 16 = 13
x/12 = 13 - 16
x/12 = -3
x = -3 * 12
x = -36
Hence the solution to the system of equation is expressed as (-36, -192)
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