document.write( "Question 102566: Use the elimination method to find all solutions of the system;\r
\n" ); document.write( "\n" ); document.write( "A. 5x+2y=-19
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\n" ); document.write( "\n" ); document.write( "b. x+3y=5
\n" ); document.write( " 6y+z=12
\n" ); document.write( " x-2x=10\r
\n" ); document.write( "\n" ); document.write( "c. 2/x+3y=16
\n" ); document.write( " -1/x+2/y=6\r
\n" ); document.write( "\n" ); document.write( "d. y=64-x^2
\n" ); document.write( " y=x^2-64
\n" ); document.write( " The two solutions of the system are:
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\n" ); document.write( "\n" ); document.write( "I have tried working the elimination rules but am confused with getting rid of x in the second equation to solve for y? \n" ); document.write( "
Algebra.Com's Answer #74590 by rmromero(383)\"\" \"About 
You can put this solution on YOUR website!

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document.write( "Use the elimination method to find all solutions of the system;\r
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document.write( "A. 5x+2y=-19
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document.write( "   7x+3y=-27
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document.write( "   x=___  y=____\r
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document.write( "    Let us eliminate x to solve for y. the equations are:\r
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document.write( "       5x + 2y = -19    eqn 1
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document.write( "       7x + 3y = -27    eqn 2\r
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document.write( "    To eliminate x, multiply -7 to eqn 1 and 5 to eqn 2. \r
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document.write( "      -7 (5x + 2y = -19)
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document.write( "       5 (7x + 3y = - 27)\r
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document.write( "       -35x - 14y =  133   eqn 1
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document.write( "        35x + 15y = -135   eqn 2
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document.write( "    Add eqn 1 and eqn 2, answer is:\r
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document.write( "       -35x - 14y =  133   eqn 1
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document.write( "        35x + 15y = -135   eqn 2
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document.write( "         0  -   y = -2\r
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document.write( "    Therefore y = -2. Substitute y = -2 to eqn 1 or eqn 2 to find x.\r
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document.write( "           5x + 2y = -19, y = -2
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document.write( "        5x + 2(-2) = -19
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document.write( "            5x - 4 = - 19          Add 4 both sides.
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document.write( "        5x - 4 + 4 = -19 + 4 
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document.write( "                5x = -15           Divide both sides by 5 so x will be left.
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document.write( "                 x = -3 \r
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document.write( "Therefore x = -3 and y = -2\r
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document.write( " You try b - d. 
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document.write( "b. x+3y=5
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document.write( "   6y+z=12
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document.write( "   x-2x=10\r
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document.write( "   Find the simpliest eqn first. look at x - 2x = 10. with this you can 
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document.write( "   solve for x.\r
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document.write( "     x - 2x = 10
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document.write( "         -x = 10    divide -1 both sides to make x positive.
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document.write( "          x = -10\r
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document.write( "   Since you have the value of x, you can substitute it to the eqn x + 3y = 5
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document.write( "     
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document.write( "         x + 3y = 5, where x = -10
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document.write( "       -10 + 3y = 5                 Add 10 both sides
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document.write( "  -10 + 10 + 3y = 5 + 10
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document.write( "             3y = 15                Divide both sides by 3
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document.write( "              y = 5\r
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document.write( "   Now you can solve for z using y = 5 and the eqn 6y+z=12.
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document.write( "         6y + z = 12 , where y = 5
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document.write( "       6(5) + z = 12
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document.write( "         30 + z = 12                Subtract both sides by 30
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document.write( "    30 - 30 + z = 12 - 30
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document.write( "              z = -18\r
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document.write( "   Therefore x = -10, y = 5 and z = -18\r
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