Solve for y= 2/3x + 1 2x + 3y = 27 using elimination.
So we have: \[y=\frac{ 2 }{ 3 }x+1\] \[2x+3y=27\] This is what we call a "system of equations" and there are various methods to solve these, we're asked to use the "elimination" one. In order to use that method, we will leave the variables on one side of the equality, and the constants on the other: \[\frac{ 2 }{ 3 }x-y=-1\] \[2x+3y=27\] Now, looking at those two equations, we must make them have opposite variables, likte this; I'll multiply the first equation by 3: \[2x-3y=-3\] \[2x+3y=27\] Now, we'll sum both equations to get: \[4x=24\] Solving for x, we get that x=6 Let's take any of the initial equations and replace the "x": \[2(6)+3y=27\] simplifying and operating, substracting 12 to each side of the equation we get: \[3y=15\] Solving for y, we get that y=5 so the solution is: \[x=6\] \[y=5\]
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