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Umnica [9.8K]
3 years ago
10

Solving systems of equations X-2y=0 Y=2x-3

Mathematics
2 answers:
bearhunter [10]3 years ago
5 0
\left \{ {{x-2y=0} \atop {y=2x-3}} \right. \\\\Substitute\ second\ equation\ to\ first:\\\\x-2(2x-3)=0\\x-4x+6=0\\-3x=-6\ \ \ \ \ |:-3\\x=2\\\\y=2\cdot2-3\\y=4-3=1\\\\ \left \{ {{y=1} \atop {x=2}} \right.
Musya8 [376]3 years ago
5 0
X - 2y = 0
y = 2x - 3

You subsitute.  Since, y = 2x - 3, then you fill what y equals in for the y on the top equation.

<em>x - 2 (2x - 3) = 0</em>     :Distribute

<em>x - 4x + 6 = 0</em><em> </em>         :Combine Like Terms

<em>-3x + 6 = 0</em>              :Subtract the 6 over to the 0
       <u>-6 </u>     <u>-6</u>
<em />
<u>(</u><em>-3x = -6</em><u>)</u> <u>÷</u> <u>-3</u>          :Now, divide the whole equation by -3

<em>x = 2</em>

Then, you want to get y, so you fill in the x that you got before into the bottom equation.

<em>y = 2 ( 2) - 3</em>           :Now, multiply 2 × 2 (Because of PEMDAS)

y = 4 - 3

<em>y = 1

</em>
<em />Your final answer is:
<em>
<u>(2, 1)</u>

</em>
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