SOLUTION: How do I solve for both methods? directions:Slove the linear equations using both substitution and combination method. y=4x-6 3y=7-3x

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Question 29875: How do I solve for both methods?
directions:Slove the linear equations using both substitution and combination method.
y=4x-6
3y=7-3x
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+y=4x-6+ and +3y=7-3x+ Substitute 4x-6 from the firt equation into the second.
+3y=7-3x+
+3%284x-6%29=7-3x+ distribute the 3
+12x-18+=+7-3x+ add 3x to both sides
+15x+-18+=+7+ add 18 to both sides
+15x+=+25+ divid both sides by 15
+x+=+25%2F15+ reduce
+x+=+5%2F3
plug x = 5/3 back into the first equation
+y=4x-6+
+y=4%285%2F3%29-6+
+y+=+%2820%2F3%29+-+6+ change 6 to 18/3 (common denom)
+y+=+%2820%2F3%29+-+18%2F3+ subtract
+y+=+2%2F3+
solution : (5/3, 2/3)
now by elimination:
+y=4x-6+ and +3y=7-3x+ we will deal with the first equation
solve for the form Ax+By=C
+y=4x-6+ subtract 4x from both sides
+-4x%2By=-6+ multiply by -1 to make the x value positive
+4x-y=6+ now solve the second equation so that it is in the same form
+3y=7-3x+ add 3x to both sides
+3x%2B3y=7+ now lets look at both equations stacked on ontop of the other.
4x - 1y = 6
3x + 3y = 7
if we multiply all parts of the first equation by three we get ...
12x - 3y = 18
3x + 3y = 7
looking at the y values .. they will cancel when we add the two equations and get ...
15x + 0y = 25 the y is gone now ...
15x = 25 divide both sides by 15
x = 25/15 or 5/3
substitute this back into ANY of the equations to find y