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`-x + 3y = 17, 4x + 3y = 7` Use any method to solve the system.

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EQ1:  `-x+3y=17`

EQ2:  `4x+3y=7`

To solve this system of equation, let's apply substitution method. To do so,  isolate the x in the first equation.

`-x+3y=17`

`-x=17-3y`

`x=(17-3y)/(-1)`

`x=-17 + 3y`

Then, plug-in this to the second equation. 

`4x+3y=7`

`4(-17+3y)+3y=7`

And, solve for y.

`-68+15y=7`

`15y=7+68`

`15y=75`

`y=75/15`

`y=5`

Now that the value of y is known, solve for x. Plug-in y = 5 to the first equation.

`-x+3y=17`

`-x+3(5)=17`

`-x+15=17`

`-x=17-15`

`-x=2`

`x=2/(-1)`

`x=-2`

Therefore, the solution is (-2,5).

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loves2learn | Student

Multiply first equation by 4.

 

After multiplying we have the following system:

`-4x+12y=6`

`4x+3y=7`

 

add the two equations together to eliminate x from the system.

`15y=75`

 

find y

`y=5`

 

substitute the value for y into the original equation to solve for x.

`-x+3(5)=17`

`x=-2`