Question 176521This question is from textbook Algebra 1 (Holt California)
: We are looking for the equations used for this problem. My son missed a day of school and missed the lesson...
This question is from textbook Algebra 1 (Holt California)
Answer by cartert(27)   (Show Source):
You can put this solution on YOUR website! The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations. If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.
x + 6 = 11
–6 –6
x = 5
You'll do something similar with the addition method.
Solve the following system using addition.
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:
2x + y = 9
3x – y = 16
5x = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:
2(5) + y = 9
10 + y = 9
y = –1
Then the solution is (x, y) = (5, –1).
It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:
3(5) – y = 16
15 – y = 16
–y = 1
y = –1
...which is the same result as before.
Solve the following system using addition.
x – 2y = –9
x + 3y = 16
Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)
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