SOLUTION: How do I Solve the following using elimination method? I get lost when I try. The whole method confesses me I don't know where to start. Math is my weakness
4x-2y=16
5x+7y=1
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-> SOLUTION: How do I Solve the following using elimination method? I get lost when I try. The whole method confesses me I don't know where to start. Math is my weakness
4x-2y=16
5x+7y=1
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Question 325893: How do I Solve the following using elimination method? I get lost when I try. The whole method confesses me I don't know where to start. Math is my weakness
4x-2y=16
5x+7y=1 Found 2 solutions by mananth, Apathious:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! 4x-2y=16...........1
5x+7y=1............2
..
you have to eliminate x or y.
to do that they have to have same co-efficients so that they cancel off.
..
How can we make the co-efficients same.
In our case lets try to eliminate x and make its co-efficients same.
..
Multiply equation 1 by 5 and equation 2 by 4.
so you get
20x-10y=80
20x+28y=4.
..
The x co-efficients are same now.
subtract the equations
(20x-10y)-(20x+28y) = (80-4)
20x-10y-20x-28y= 76
-38y= 76
Divide by -38
y= 76/-38
y= -2
..
Plug the value of 2 in the equation 1
4x-2y=16
4x-2*-2=16
4x+4 =16
add -4
4x+4-4 = 16-4
4x=12
divide by 4
4x/4 = 20/4
x=5
x= 5 & y=-2
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You can put this solution on YOUR website! 4x-2y=16_5x+7y=1
Multiply each equation by the value that makes the coefficients of y equal. This value is found by dividing the least common multiple of the coefficients of y by the current coefficient. In this case, the least common multiple is 14.
7*(4x-2y=16)_2*(5x+7y=1)
Multiply each equation by the value that makes the coefficients of y equal. This value is found by dividing the least common multiple of the coefficients of y by the current coefficient. In this case, the least common multiple is 14.
7*(4x-2y)=7(16)_2*(5x+7y)=2(1)
Multiply 7 by each term inside the parentheses.
7*(4x-2y)=112_2*(5x+7y)=2(1)
Multiply 7 by each term inside the parentheses.
(28x-14y)=112_2*(5x+7y)=2(1)
Remove the parentheses around the expression 28x-14y.
28x-14y=112_2*(5x+7y)=2(1)
Multiply 2 by each term inside the parentheses.
28x-14y=112_2*(5x+7y)=2
Multiply 2 by each term inside the parentheses.
28x-14y=112_(10x+14y)=2
Remove the parentheses around the expression 10x+14y.
28x-14y=112_10x+14y=2
Add the two equations together to eliminate y from the system.
10x+14y=2_28x-14y=112_38x =114
Divide each term in the equation by 38.
x=3
Substitute the value found for x into the original equation to solve for y.
28(3)-14y=112
Multiply 28 by each term inside the parentheses.
84-14y=112
Move all terms not containing y to the right-hand side of the equation.
-14y=28
Divide each term in the equation by -14.
y=-2
This is the final solution to the independent system of equations.
x=3_y=-2
