



 
Hi, I think there is a typo in your question. I think the first + in the second equation should be =. Let me know if I am mistaken. If I am correct then the equations are \[2x = 3y \mbox{ and } x2 = \frac43 (y+1).\] The first task I would perform is to eliminate the fraction in the second equation. To do this multiply both sides of the second equation by 3 to get \[3\left(x2\right) = 3\left(\frac43 (y+1)\right)\] which then becomes \[3x  6 = 4(y + 1).\] Simplify and then solve the simplified equation with the first equation. Make sure you verify your answer. Write back if you need more assistance, 



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