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.

Equations

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.

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.