How do you solve this system of equations: $6 x - 10 y = - 20 and 5 x + 3 y = 1$ $6x - 10y = - 20 and 5x + 3y = 1$ ?

Question

2088 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-20T10:11:51

$(- \frac{25}{34}, \frac{53}{34})$ $(-25/34,53/34)$

$6 x - 10 y = - 20$ $6x-10y=-20$
$5 x + 3 y = 1$ $5x+3y=1$

Elimination would be the best option inside of substitution since you're going to end up with fractions and fractions are annoying.

Either way, we're still going to be working with them but using the elimination method will allow us to use fractions once. You'll see.

Let's say we'll be getting rid of $x$ $x$ , so we'll multiply the top equation by $5$ $5$ and the bottom by $6$ $6$ .

$30 x - 50 y = - 100$ $30x-50y=-100$
$30 x + 18 y = 6$ $30x+18y=6$

Then, subtract the two equations so we end up with one. After that, solve for $y$ $y$ .

$- 68 y = - 106$ $-68y=-106$

$y = \frac{53}{34}$ $y=53/34$

Sub $\frac{53}{34}$ $53/34$ for $y$ $y$ in one of the equations and solve for $x$ $x$ .

$5 x + 3 \cdot \frac{53}{34} = 1$ $5x+3*53/34=1$
$5 x = - \frac{125}{34}$ $5x=-125/34$
$x = - \frac{25}{34}$ $x=-25/34$

Thus, we get the answer:

$(- \frac{25}{34}, \frac{53}{34})$ $(-25/34,53/34)$

Noice

How do you solve this system of equations: 6x−10y=−20and5x+3y=16x - 10y = - 20 and 5x + 3y = 1?