Math Question - Solve Equation for Two Unknowns
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Math Question - Solve Equation for Two Unknowns











































Solve the following system of equations. If there are no solutions, type "No Solution" for both x and y. If there are infinitely many solutions, type "x" for x, and an expression in terms of x for y.

1x−3y=2
2x−2y=5

x = .
y = .

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Mar 24, 2011
Solve Equation for Two Unknowns
by: Staff


The question:

Solve the following system of equations. If there are no solutions, type "No Solution" for both x and y. If there are infinitely many solutions, type "x" for x, and an expression in terms of x for y.

1x−3y=2
2x−2y=5

x = .
y = .


The answer:

The fastest way to eliminate one of the variables is to add the two equations together.

1x − 3y = 2
2x − 2y = 5

Multiply each side of the first equation by (-2):

(-2)*(1x − 3y) = (-2)*(2)
2x − 2y = 5

Eliminate the parentheses in the first equation using the distributive law:

(-2)*(1x) − (-2)*(3y) = (-2)*(2)
2x − 2y = 5

-2x + 6y = -4
2x − 2y = 5

Add the two equations. Notice that the variable x is eliminated in the process. This will allow us to solve for the value of y:

-2x + 6y = -4
2x − 2y = 5
-----------------
-2x + 2x + 6y - 2y = -4 + 5

0 + 6y - 2y = -4 + 5

6y - 2y = -4 + 5

4y = -4 + 5

4y = 1

Divide each side of the equation by 4:

4y = 1

4y/4 = 1/4

y*(4/4) = 1/4

y*(1) = 1/4

y = 1/4

Since we now know the numerical value of y, we can substitute this value into either of the two original equations, and then solve for x.

1x − 3y = 2

x − 3*(1/4) = 2

x − .75 = 2

add .75 to each side of the equation:

x − .75 + .75 = 2 + .75

x + 0 = 2 + .75

x = 2 + .75

x = 2.75

The final answer is:

x = 2.75

y = .25


Check the answer by substituting the numerical values of x and y into the original equations:

1x − 3y = 2

2.75 – 3*.25 = 2, correct

2x − 2y = 5

2*2.75 − 2*.25 = 5, correct

Since the numerical values of x and y work in both of the original equations, the solutions are correct.




Thanks for writing.


Staff
www.solving-math-problems.com


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