SOLUTION: Solve by elimination method:
(5/2) x + (10/3)y = 55 and (7/3)x + (7/2)y = 56
I know these two equations then equal:
15x+20y=330 and 14x+21y=336
But I don't know what to do fr
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Question 1028377: Solve by elimination method:
(5/2) x + (10/3)y = 55 and (7/3)x + (7/2)y = 56
I know these two equations then equal:
15x+20y=330 and 14x+21y=336
But I don't know what to do from there! And what is elimination method anyway? What's the difference between that and substitution method or are they the same thing and are there any other kinds of methods?
Found 3 solutions by mananth, josgarithmetic, Shin123:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
15x+20y=330
/5
3x+4y=66
14x+21y=336
/7
2x+3y=48
3 x + 4 y = 66 .............1
2 x + 3 y = 48 .............2
Eliminate y
multiply (1)by -3
Multiply (2) by 4
-9 x -12 y = -198
8 x + 12 y = 192
Add the two equations
-1 x = -6
/ -1
x = 6
plug value of x in (1)
3 x + 4 y = 66
18 + 4 y = 66
4 y = 66 -18.00
4 y = 48
y = 12
x= 6
y= 12
m.ananth@hotmail.ca
Answer by josgarithmetic(39621) (Show Source): You can put this solution on YOUR website!
Elimination is just a fancy often more efficient way of handling Substitution.
Multiply each equation so you can eliminate y and solve for x; start over and multiply each equation to eliminate x and solve for y.
System starts as
You can simplify each of the equations before you start the elimination method process because each of your equations is still simplifiable. The factor 5 in one equation and factor 7 in the other.
You will want coefficient on y to become 12, which I will not explain...
, the system is equivalent to the first system (and to the second system).
This next step is two substitutions, USING the system...
.
.
...and what is x ?
Now, you can manage the process again to find y in similar manner.
Answer by Shin123(626) (Show Source): You can put this solution on YOUR website!
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