# Q31 MET115 Solving the System | Summary and Q&A

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September 27, 2014
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Q31 MET115 Solving the System

## TL;DR

Learn how to solve equations using the elimination method by eliminating one variable to find the value of the other variable.

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### Q: How can we solve the given equations?

The elimination method is used in this video. By multiplying one equation to match the coefficients of one variable, we can add the equations and eliminate that variable.

### Q: What is the value of y?

Dividing both sides of the equation 28y = 5.6 by 28 gives y = 20.

### Q: What is the value of x?

Substituting the value of y (20) back into the first equation x + y = 70, we can solve for x. Subtracting 20 from both sides gives x = 50.

### Q: How can we ensure we choose the correct answer?

Following the solution, we can determine that the amount of milliliters (x) corresponds to the 32% solution, while the value of y corresponds to the 60% solution. Therefore, we should choose Choice C.

## Summary & Key Takeaways

• Two equations are given: x + y = 70 and 32x + 6y = 288.

• The elimination method is used by multiplying the first equation by -32 to match the coefficients of x in both equations.

• After adding the two equations together, the variable x is eliminated, leaving 28y = 5.6.

• Dividing both sides by 28 gives y = 20, and substituting this value back into the first equation gives x = 50.

• Choice C is the correct answer, representing the solution for the given equations.