Question

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.

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asked 2021-03-10

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

Answers (1)

2021-03-12

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

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