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

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

Nancy Johnson · Accepted Answer

Step 1  Consider the system of equation  −6A+8B=48.....(1)    −6A+2B=42.....(2)  Subtract equation (2) from equation (1)  −6A+8B−(−6A+2B)=48−42  −6A+8B+6A−2B=48−42  6B=6 (Combine the like term)  Divide both sides by 6 and further simplify it  6B6=66  B=1  Step 2  Substitute B=1 into −6A+8B=48  −6A+8×1=48  −6A+8=48  Subtract 8 from sides and further simplify it  −6A+8−8=48−8  −6A=40  Divide both sides by -6 and further simplify it  −6A−6=40−6  A=−203  Therefore, A=−203 and B=1  Hence, the given system of equation has only one solution.

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

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