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# elimination method

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Solve system of two equations with x and y (use any technique):
1. 2x - y = 0
3x + 2y = 7

2. x + 2y = 4
2x - y = -7

3. 4x + y = 15
2x + 5y = 21

Assign x and y for unknown values, create system of two equations and find solution:

4. Investments. Michael Perez has a total of \$2000 on deposit with two savings institutions. One pays interest at the rate of 6%/year, whereas the other pays interest at the rate of 8%/year. If Michael earned a total of \$144 in interest during a single year, how much does he have on deposit in each institution?

5. The Coffee Shoppe sells a coffee blend made from two coffees, one costing \$5/lb and the other costing \$6/lb. If the blended coffee sells for \$5.60/lb, find how much of each coffee is used to obtain the desired bend. Assume that the weight of the blended coffee is 100lb.

6. Simple calculator costs \$5 and scientific calculator costs \$16. The store sold 35 calculators and received \$340.
How many of each calculator was sold?

https://brainmass.com/math/discrete-math/mathematics-solving-systems-equations-371091

#### Solution Preview

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Solve system of two equations with x and y (use any technique):
1. 2x - y = 0
3x + 2y = 7

The process I prefer is the elimination method in which we must multiply one equation by a constant so that if we add the two equations we are left with only one variable. So in this instance we will multiply the top equation by 2. This way when we add the two equations -2y + 2y = 0 and we are left with only one variable:

2(2x - y = 0)
3x + 2y = 7

4x - 2y = 0
+ 3x + 2y = 7
7x + 0 = 7

now just solve for x:

7x = 7
x = 1

Using this value for x we will substitute it into either one of our original equations to find our value for y:

2(1) - y = 0
2 - y = 0
2 = y

Thus our solution is (1, 2)

2. x + 2y = 4
2x - y = -7

We will use the same process here as well this time we will multiply the second equation by 2:

x + 2y = 4
2(2x - y = -7)

x + 2y = 4
+ 4x - 2y = -14
...

#### Solution Summary

This solution helps with solving systems of equations.

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