How do you find the point of intersection for $x + 2 y = 3$ $x+2y=3$ and $3 x - y = - 5$ $3x-y= -5$ ?

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Answer 1 · 2015-07-08T20:10:00

You use one equation to express one of the variables in terms of the other. Then you substitute that in the other equation.

$3 x - y = - 5 \to - y = - 3 x - 5 \to y = 3 x + 5$ $3x-y=-5->-y=-3x-5->y=3x+5$

Then:
$x + 2 \cdot (3 x + 5) = 3 \to 7 x + 10 = 3 \to x = - 1$ $x+2*(3x+5)=3->7x+10=3->x=-1$

From the first equation:
$y = 3 x + 5 = 3 \cdot (- 1) + 5 = 2$ $y=3x+5=3*(-1)+5=2$

So the intersection point is $(- 1, 2)$ $(-1,2)$

Check your answer with the original equations!
$- 1 + 2 \cdot 2 = 3 and 3 \cdot (- 1) - 2 = - 5$ $-1+2*2=3and3*(-1)-2=-5$

How do you find the point of intersection for x+2y=3x+2y=3x+2y=3 and 3x-y= -53x−y=−53x-y= -5?