SOLUTION: 5. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the
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-> SOLUTION: 5. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the
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Question 40431This question is from textbook INTERMEDIATE ALGEBRA
: 5. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
6x + 2y = 2
3x + 5y = 5 This question is from textbook INTERMEDIATE ALGEBRA
You can put this solution on YOUR website! Solve this system of equations using the elimination method.
6x + 2y = 2
3x + 5y = 5
First we need to decide which coefficiant to work with in order to cancel out or "eliminate" the other.
For instance, we have 6x and 3x.
What do we need to do to the 3x in order to cancel out the 6x?
What if we multiply the second (bottom) equation by -2.
6x + 2y = 2
-2(3x + 5y = 5)
When we distribute, we get...
6x + 2y = 2
-6x -10y = -10
Now we can add our system and eliminate the x (for now) and solve the equation for y. Like so...
6x + 2y = 2
-6x -10y = -10
_______________
0 - 8y = -8
or...
-8y = -8
We can now divide both sides by -8...
-8/-8 = 1
So...
y = 1
But wait, we're not done yet!
Now we have to insert the value for y into one of the original equations in order to get the value for x.
So, let's take...
6x + 2y = 2
Insert the value for y = 1.
6x + 2(1) = 2
Now solve for x...
6x + 2 = 2
6x = 0
if we divide both sodes by 6 we get 0/6 which equals 0
So...
x = 0
Now we should check to see if our answers make the original equations true
6x + 2y = 2
3x + 5y = 5
or...
6(0) + 2(1) = 2
3(0) + 5(1) = 5
These are both true.
So our solution set is...
(0,1)
I hope this helps
Good Luck!
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