SOLUTION: Please solve each system of equations. x+y=8 x+3y=14 Thank you!

Question 36870This question is from textbook Holt Pre Algebra
: Please solve each system of equations.
x+y=8
x+3y=14
Thank you! This question is from textbook Holt Pre Algebra

Answer by martinr(6) (Show Source): You can put this solution on YOUR website!
First off, there are three ways to solve this. One by graphing, the least accurate method, but the one everyone is most familiar with.
The other two are solved algebraically.
I like the elimination method since you did not indicate which method to use.
In the elimination method, you need to have the problem set up with the like terms in columns, just like the problem is written. Next, choose one of the variables to eliminate, with this method you will need to make oppoistes for either the x or the y varables. Once again it makes no difference which variable you choose to eliminate. To make opposites, you need to multiply one of both of the equations by a number, and sometimes , the problem will already have opposites given.
x+y=8
x+3y=14 I am going to eliminate the y variable. The opposite of 3y is -3y.
In order to eliminate the y variable, I will multiply the top equation by -3.
(-3) (x + y = 8) -3x - 3y = -24
(x + 3y = 14) (x + 3y = 14)
Next add the columns.
-3x + x = -2x
-3y + 3y = 0
-24 = 14 = -10
Now I have
-2x = -10 Divide by -2
x = 5
This is only half the solution, we need the y value. You can once again use either original equation.
I will use the top one, x + y = 8
x + y = 8, we alreadt know that x = 5
5 + y = 8, subtract 5 from each side
y = 3
Therefore the solution is (5, 3).
You can check your solution by plugging both the x and y values into the original equation. It needs to satisfy both eqautions in order to be a solution. If one fails, either you check is incorrect, or you have the incorrect solution.
Good Luck
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