3
$\begingroup$

How do I find the intersection of these two lines with their equations in general form. I don't want to graph them and I'm wondering if its possible with out converting them to gradient intercept form?

$2x+3y-5=0$ and $5x-y-4=0$

Thank you!

2 Answers 2

1

From the second equation, we have that $y=5x-4$. Plugging that into the first equation we get $0 = 2x+3y - 5 = 2x + 3(5x-4) -5 = 2x + 15x - 12 - 5 = 17x-17.$ Solving for $x$, and then plugging into $y=5x-4$, gives the point of intersection.

  • 0
    @Eugene: No, you're right.2012-06-10
1

We will label the two equations: $ 2x +3y = 5 \qquad (1) $ and $ 5x - y = 4 \qquad (2) $

Multiply the second equation by $3$ to get $ 15 x -3y = 12 \qquad (3) $

By adding $(3)$ to $(1)$ we get $ 17x = 17. $ Therefore $x = 1$. Now plugging in $x = 1$ to $(2)$ we get $ 5 -y = 4 $ which means $ - y = -1 $ and so $y = 1$.

We check and find that it is indeed the case ie $ 2(1) + 3(1) = 5 $ and $ 5(1) - 1 = 4. $