SOLUTION: solve by using substitution or elimination x=3y+11 2x+5y=0

Question 124738: solve by using substitution or elimination
x=3y+11
2x+5y=0
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Substitution is an easy way to solve this problem. What makes it easy is the fact that the top
equation is already solved for one variable (x) in terms of the other variable. It tells you
that x = 3y +11. You can use this information by going to the second equation and in it you
replace x by its equal 3y + 11.
.
The second equation is:
.
2x + 5y = 0
.
If you replace x by 3y + 11 this second equation becomes:
.
2(3y + 11) + 5y = 0
.
Do the distributed multiplication (multiply 2 times each of the terms in the parentheses)
and the equation becomes:
.
6y + 22 + 5y = 0
.
Add the like terms 6y and 5y to reduce the left side of this equation to:
.
11y + 22 = 0
.
Subtract 22 from both sides to get rid of the 22 on the left side. This subtraction
results in the equation becoming:
.
11y = -22
.
Solve for y by dividing both sides of this equation by 11 (the multiplier of x) and the
result is:
.
y = -22/11 = -2
.
So now you know one of the answers ... y is -2. You can then solve for x by going back
to either of the two original equations, substituting -2 for y in that equation, and solving
for x. The easiest equation to return to is the equation:
.
x = 3y + 11
.
Substitute -2 for y and this equation becomes:
.
x = 3*(-2) + 11
.
and this reduces to:
.
x = -6 + 11 = +5
.
Now you know the two variables ... x equals 5 and y equals -2.
You can check your answers by substituting them into the two original equations and verifying
that the left side of the equation does equal the right side.
.
First check:
.
x = 3y + 11
.
5 = 3(-2) + 11
.
5 = -6 + 11
.
5 = 5
.
That checks.
.
Second check:
.
2x + 5y = 0
.
2(5) + 5(-2) = 0
.
10 -10 = 0
.
0 = 0
.
That also checks.
.
Since the answers for x and y satisfy both of the original equations, these answers are
correct.
.
Hope this helps you to understand the process of substitution in solving a pair of independent
linear equations.
.
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