Use addition to solve the linear system of equations. Include all of your work in your final answer.

1 answer:

4 0

We have that
<span>x - y = 4-------> multiply by -1-----> -x+y=-4-----> equation 1
x+ 2y = 4------> equation 2

add equation 1 and equation 2
-x+y=-4
x+2y=4
-----------------
0+3y=0-----> y=0
x+2y=4------> x+0=4------>x=4

the solution is the point (4,0)</span>

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<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20%7D%7Bxy%7D%20%20%5Ctimes%20%20%5Cfrac%7B%20%7B4xy%7D%5E%

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$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------\\\\
\cfrac{3x^2}{xy}\cdot \cfrac{4xy^2}{\frac{1}{y}}\implies \cfrac{3x^2}{xy}\cdot \cfrac{4xy^2}{y^{-1}}\implies \cfrac{3x^2x^1y^2}{xy^1y^{-1}}\implies \cfrac{3x^{2+1}y^2}{xy^{1-1}}
\\\\\\
\cfrac{3x^3y^2}{xy^0}\implies \cfrac{3x^3y^2}{x^1}\implies 3x^3y^2x^{-1}\implies 3x^{3-1}y^2\implies 3x^2y^2$