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!
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!
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.
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 $$ 5(1) - 1 = 4. $$