document.write( "Question 58649This question is from textbook
\n" ); document.write( ": Solve the systems of equations graphically
\n" ); document.write( "{4x-3y=-4
\n" ); document.write( "{2x-5y=-4
\n" ); document.write( "

Algebra.Com's Answer #40266 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
In order to solve this graphically we have to plot the graphs.
\n" ); document.write( "Arrange the equation in the \"y=\" format\r
\n" ); document.write( "\n" ); document.write( "4x - 3y = -4
\n" ); document.write( "-3y = -4 - 4x
\n" ); document.write( "3y = 4x + 4, multiplied equation by -1, y should be positive
\n" ); document.write( "y = (4/3)x + (4/3)
\n" ); document.write( ":
\n" ); document.write( "Find the value of y by substituting for x:
\n" ); document.write( " x | y
\n" ); document.write( "-2 | -(4/3)
\n" ); document.write( "-1 | 0
\n" ); document.write( "0 | +(4/3)
\n" ); document.write( "+1 | +(8/3)
\n" ); document.write( "+2 | +4
\n" ); document.write( "Plot these on a graph: x axis 4/-4 and y axis 4/-4 will work ok
\n" ); document.write( ":
\n" ); document.write( "Do the same with the 2nd equation:\r
\n" ); document.write( "\n" ); document.write( "2x-5y=-4
\n" ); document.write( "-5y = -4 - 2x
\n" ); document.write( "5y = 2x + 4
\n" ); document.write( "y = (2/5)x + (4/5)
\n" ); document.write( ":
\n" ); document.write( "Find the value of y by substituting for x:
\n" ); document.write( " x | y
\n" ); document.write( "-2 | 0
\n" ); document.write( "-1 | +(2/5)
\n" ); document.write( "0 | +(4/5)
\n" ); document.write( "+1 | +(6/5)
\n" ); document.write( "+2 | +(8/5)
\n" ); document.write( "Plot this on the same graph, it should look like this(green is 2nd equation):\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( ":
\n" ); document.write( "The solution is the x,y values where the lines cross. YOu can see it is not
\n" ); document.write( "going to be an integer. I solved using elimination, got x = -(4/7) y = +(4/7)
\n" ); document.write( "which looks about right
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