`-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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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`
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