`(x - 1)/2 + (y + 2)/3 = 4, x - 2y = 5` Solve the system by the method of substitution.

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EQ1:   `(x -1)/2+(y+2)/3 =4`

EQ2:  `x-2y=5`

To solve using method of substitution,  we have to isolate one of the variable. 

For this system of equations, it is better that we isolate the x in the second equation.

`x-2y=5`

`x=5+2y`

Then, plug-in this to the first equation.

`(x-1)/2+(y+2)/3=4`

`(5+2y-1)/2+(y+2)/3=4`

And solve for y.

`(2y+4)/2+(y+2)/3=4`

To solve this, it is better to eliminate the fractions in our equation. This can be done by multiplying both sides by the LCD. The LCD of the fractions present is 6.

`6*((2y+4)/2+(y+2)/3)=4*6`

`3(2y+4) + 2(y+2)=24`

`6y+12+2y+4=24`

`8y+16=24`

`8y=24-16`

`8y=8`

`y=8/8`

`y=1`

Now that the value of y is known, solve for x. Plug-in y=1 to the second equation.

`x -2y=5`

`x-2(1)=5`

`x-2=5`

`x=5+2`

`x=7`

Therefore, the solution is (7,1).

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