 # A system of linear equations is given below.2x+4y=10-\frac{1}{2}x+3=yFind the solution to the system of equations.A. (0, -3)B. (-6, 0)C. Chaya Galloway 2021-03-10 Answered

A system of linear equations is given below.
$2x+4y=10$
$-\frac{1}{2}x+3=y$
Find the solution to the system of equations.
A. (0, -3)
B. (-6, 0)
C. There are infinite solutions.
D. There are no solutions.

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Step 1
Given system of equations are:
$2x+4y=10$
$-\frac{1}{2}x+3=y$
Solution:
We have equations:
$2x+4y=10...\left(1\right)$
$-\frac{1}{2}x+3=y$
$-x+6=2y$
$x+2y=6...\left(2\right)$
Step 2
We know that if:
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne q\frac{{c}_{1}}{{c}_{2}}$ then two lines are parallel.
And when two lines are parallel then there is no solution.
We have, .
$\frac{{a}_{1}}{{a}_{2}}=\frac{2}{1}=2$
$\frac{{b}_{1}}{{b}_{2}}=\frac{2}{1}=2$
$\frac{{c}_{1}}{{c}_{2}}=\frac{10}{6}=\frac{5}{3}$
We find that $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne q\frac{{c}_{1}}{{c}_{2}}$.
Step 3
Therefore, given two lines are parallel.
Therefore, there are no solutions.
Hence, option (D) is correct answer.