document.write( "Question 220150: solve trhe following system of equations using substitution.\r
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document.write( "2x + 3y = -1
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document.write( "5x - 2y = -12\r
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Algebra.Com's Answer #165349 by jsmallt9(3758)![]() ![]() ![]() You can put this solution on YOUR website! (Note: The solution provided by someone else did not use the substitution method.) \n" ); document.write( "The substitution method:
\n" ); document.write( "1. Solve an equation for one of the variables. The \"easy\" systems for the substitution method are the ones where a variable has a coefficient of 1 (or -1). Unfortunately this system is not an \"easy\" one. Since I prefer positives I'll solve the second equation for y: \n" ); document.write( "5x - 2y = -12 \n" ); document.write( "Add 2y to both sides: \n" ); document.write( "5x = 2y - 12 \n" ); document.write( "Add 12 to both side: \n" ); document.write( "5x + 12 = 2y \n" ); document.write( "Multiply both sides by 1/2 (or divide by two): \n" ); document.write( " \n" ); document.write( "2. Substitute into the other equation: \n" ); document.write( "2x + 3y = -1 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "3. Solve this equation. \n" ); document.write( "Simplify. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Subtract 18 from each side: \n" ); document.write( " \n" ); document.write( "Multiply by 2/19 (or divide by 19/2): \n" ); document.write( " \n" ); document.write( "4. Substitute this solution back into one of the original equations: \n" ); document.write( "2x + 3y = -1 \n" ); document.write( "2(-2) + 3y = -1 \n" ); document.write( "5. Solve this equation. \n" ); document.write( "Simplify. \n" ); document.write( "-4 + 3y = -1 \n" ); document.write( "Add 4 to both sides: \n" ); document.write( "3y = 3 \n" ); document.write( "Multiply both sides by 1/3 (or divide by 3): \n" ); document.write( "y = 1 \n" ); document.write( "So the solution is the point (-2, 1) \n" ); document.write( " |