document.write( "Question 718316: Hello,
\n" ); document.write( "I am trying to graph an absolute value equation and cannot seem to solve it.
\n" ); document.write( "the equation is to Solve |2x+2|=-2/3x+2 Graphically.
\n" ); document.write( "In my answer booklet it says the 2 answer to this absolute value equation is x=-3 and x=0 but when I plot these two equations i notice that they both have a y intercept of 2 so, i thought that they would meet at x=2. Please help me, it would be greatly appreciated! Best of Luck.
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Algebra.Com's Answer #440954 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
First of all, please put fractions which are factors in parentheses. The right side, as you posted it, means:
\n" ); document.write( "\"-2%2F3x%2B2\"
\n" ); document.write( "It was only your mention that the y-intercept was 2 that allowed me to figure out that it was supposed to be:
\n" ); document.write( "\"%28-2%2F3%29x%2B2\"

\n" ); document.write( "description that the
At the y-intercept of 2, the y-coordinate is 2. The x-coordinate on the y-axis is always 0. This explains the x=0 solution.

\n" ); document.write( "As for the other solution...
\n" ); document.write( "The graphs of equations of the form:
\n" ); document.write( "y = | mx+b |
\n" ); document.write( "will \"bounce\" off the x-axis. They will \"bounce\" at the point whose x-coordinate is the number that makes mx+b be zero. For y = | 2x+2 |, the graph will \"bounce\" off the x-axis at -1 because -1 makes 2x+2 be zero. (If you can't see this, just set 2x+2 = 0 and solve for x.)

\n" ); document.write( "The graph of y = | mx + b | will be the graphs of two \"half-lines\" which meet at the \"bounce\" point. One \"half-line\" will the part of the graph of y = mx+b that is above the x-axis and the other \"half-line\" will the the part of the graph of y = -(mx+b) (or y - -mx-b) which is above the x-axis.

\n" ); document.write( "For y = | 2x+2 |, the two \"half-lines\" would be: y = 2x+2 and y = -2x-2. The first \"half-line\" is the one whose y-intercept is 2 and intersects \"y+=+%28-2%2F3%29x+%2B+2\" there. The other \"half-line\", y = -2x-2, will intersect \"y+=+%28-2%2F3%29x+%2B+2\" somewhere to the left of the \"bounce\" point at (-1, 0).

\n" ); document.write( "Here's a graph of both y = | 2x+2 | and \"y+=+%28-2%2F3%29x+%2B+2\":
\n" ); document.write( "\"graph%28400%2C+400%2C+-5%2C+5%2C+-2%2C+8%2C+abs%282x%2B2%29%2C+%28-2%2F3%29x+%2B+2%29\"
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