college algebra

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please solve by method of substitution
0.3x-0.4y-0.33=0
0.1x + 0.2y -.21=0

please give all steps to solve.I think I understand the basic concept but this is little more complicated for me. Thanks in advance

• college algebra -

Things look a lot easier if you multiply each equation by 100 and simplify, and transpose the constants to the right hand side.
30x - 40y = 33 .....(1)
10x + 20y = 21 .....(2)

From equation (2), transpose 20y to the right and divide equation by 10 to get
10x = 21 - 20y
x = 2.1 -2y .... (3)
Substitute x into equation (1) to get
30(2.1-2y) - 40y = 33
63 - 60y -40y = 33
-100y = 33-63
-100y = -30
y=(-30)/(-100)
=0.3
Substitute y=0.3 into (3) to get
x=2.1-2(0.3)
=1.5

Now, substitute these values of x=1.5 and y=0.3 into the original equations to make sure the solution is correct.

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