How can one halfx βˆ’ 5 = one thirdx + 6 be set up as a system of equations?

Accepted Solution

Answer:The system of equations is[tex]x-2y=10[/tex][tex]x-3y=-18[/tex]Step-by-step explanation:we have[tex]\frac{1}{2}x-5=\frac{1}{3}x+6[/tex]Let Β set both sides of the equation separately equal to ysoLeft side[tex]y=\frac{1}{2}x-5[/tex] Multiply both sides by 2 to remove the fraction[tex]2y=x-10[/tex] Rewrite as standard equation[tex]x-2y=10[/tex] -----> equation ARight side[tex]y=\frac{1}{3}x+6[/tex] Multiply both sides by 3 to remove the fraction[tex]3y=x+18[/tex] Rewrite as standard equation[tex]x-3y=-18[/tex] -----> equation BthereforeThe system of equations is[tex]x-2y=10[/tex][tex]x-3y=-18[/tex]