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Sagot :
First, we need to find the equation for the two equations.
The equation graphed has a y-intercept of 3 and a slope of
[tex]m=\frac{-6}{3}=-3[/tex]therefore, the equation of the line is
[tex]\boxed{y=-\frac{1}{2}x+3.}[/tex]For the second equation, we know what it has a slope of 3; therefore it can be written as
[tex]y=3x+b[/tex]Now, we also know that this equation passes through the point y = -5, x = 2; therefore,
[tex]-5=3(2)+b[/tex]which gives
[tex]-5=6+b[/tex][tex]b=-11[/tex]Hence, the equation of the line is
[tex]\boxed{y=3x-11}[/tex]Now we have the equations
[tex]\begin{gathered} y=-\frac{1}{2}x+3 \\ y=3x-11 \end{gathered}[/tex]equating them gives
[tex]-\frac{1}{2}x+3=3x-11[/tex]adding 11 to both sides gives
[tex]-\frac{1}{2}x+14=3x[/tex]adding 1/2 x to both sides gives
[tex]14=\frac{7}{2}x[/tex]Finally, dividing both sides by 7/2 gives
[tex]\boxed{x=4\text{.}}[/tex]The corresponding value of y is found by substituting the above value into one of the equations
[tex]y=-\frac{1}{2}(4)+3[/tex][tex]y=1[/tex]Hence, the solution to the system is
[tex](4,1)_{}[/tex]
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