SOLUTION: .813x+.743y=.079 .491x+.114y=.826 solve by substitution, graphing, manipulating coefficients, and cramer's rule, i cant figure out what to do because all the variables have num

Question 89933This question is from textbook intermediate algebra
: .813x+.743y=.079
.491x+.114y=.826
solve by substitution, graphing, manipulating coefficients, and cramer's rule, i cant figure out what to do because all the variables have numbers????? This question is from textbook intermediate algebra

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
.813x+.743y=.079
.491x+.114y=.826
---------------
To get rid of the decimals, multiply thru each equation by 1000:
1st: 813x + 743y = 79
2nd: 491x + 114y = 826
-----------------------
To solve the system by elimination, multiply 1st thru by 491 and 2nd thru by 813
3rd: 399183x + 364813y = 38789
4th: 399183x + 92682y = 671538
--------------------------
To solve for y, subtract 4th from 3rd to get:
272131y = -632749
y = -2.325
------------
Substitute that into 491x + 114y = 826 to solve for x:
491x + 114*-2.325 = 826
491x = 826 + 265.069
491x = 1091.069
x = 2.222
============
Comment: I checked these answers with matrix methods.
They are correct.
Cheers,
Stan H.

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