147k views
2 votes
Solving systems of equations

X-2y=0
Y=2x-3

2 Answers

0 votes
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.

x - 2 (2x - 3) = 0 :Distribute

x - 4x + 6 = 0 :Combine Like Terms

-3x + 6 = 0 :Subtract the 6 over to the 0
-6 -6

(-3x = -6) ÷ -3 :Now, divide the whole equation by -3

x = 2

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

y = 2 ( 2) - 3 :Now, multiply 2 × 2 (Because of PEMDAS)

y = 4 - 3

y = 1

Your final answer is:

(2, 1)

User Joosep Parts
by
8.1k points
5 votes

\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.
User Asifa
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories