# Solve the following system. Enter the values for A and B as reduced fr

Solve the following system. Enter the values for A and B as reduced fractions or integers.
$$\displaystyle-{6}{A}+{8}{B}={48}$$
$$\displaystyle-{6}{A}+{2}{B}={42}$$
- One or more solutions:
- No solution
- Infinite number of solutions

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Nancy Johnson
Step 1
Consider the system of equation
$$\displaystyle-{6}{A}+{8}{B}={48}$$.....(1)
$$\displaystyle-{6}{A}+{2}{B}={42}$$.....(2)
Subtract equation (2) from equation (1)
$$\displaystyle-{6}{A}+{8}{B}-{\left(-{6}{A}+{2}{B}\right)}={48}-{42}$$
$$\displaystyle-{6}{A}+{8}{B}+{6}{A}-{2}{B}={48}-{42}$$
$$\displaystyle{6}{B}={6}$$ (Combine the like term)
Divide both sides by 6 and further simplify it
$$\displaystyle{\frac{{{6}{B}}}{{{6}}}}={\frac{{{6}}}{{{6}}}}$$
$$\displaystyle{B}={1}$$
Step 2
Substitute $$\displaystyle{B}={1}$$ into $$\displaystyle-{6}{A}+{8}{B}={48}$$
$$\displaystyle-{6}{A}+{8}\times{1}={48}$$
$$\displaystyle-{6}{A}+{8}={48}$$
Subtract 8 from sides and further simplify it
$$\displaystyle-{6}{A}+{8}-{8}={48}-{8}$$
$$\displaystyle-{6}{A}={40}$$
Divide both sides by -6 and further simplify it
$$\displaystyle{\frac{{-{6}{A}}}{{-{6}}}}={\frac{{{40}}}{{-{6}}}}$$
$$\displaystyle{A}=-{\frac{{{20}}}{{{3}}}}$$
Therefore, $$\displaystyle{A}=-{\frac{{{20}}}{{{3}}}}$$ and $$\displaystyle{B}={1}$$
Hence, the given system of equation has only one solution.