# How do you solve the system of equations -3x-4y=20 and x-10y=16?

$x = - 4$ and $y = - 2$

#### Explanation:

Given system of equations

$- 3 x - 4 y = 20 \setminus \ldots \ldots . . \left(1\right)$

$x - 10 y = 16 \setminus \ldots \ldots . . \left(2\right)$

Multiplying (2) by $3$ & adding to (1) as follows

$- 3 x - 4 y + 3 \left(x - 10 y\right) = 20 + 3 \setminus \cdot 16$

$- 34 y = 68$

$y = - 2$

setting $y = - 2$ in (1) we get

$x = \setminus \frac{- 4 y - 20}{3}$

$= \setminus \frac{- 4 \left(- 2\right) - 20}{3}$

$= - 4$

Jul 9, 2018

Express $x$ as a function of $y$ and replace in the second equation.

#### Explanation:

To get a very fast answer to your problem, just use one of the two equations to express $x$ or $y$. In this case, let's do it with $x$.

So, we have the following system:

1) \ -3x-4y=20
2) \ x-10y=16

If we express $x$ in 2), then we have:

$x = 16 + 10 y$

Then we replace $x$ in 1) with what we obtained with 2) and we get:

$- 3 \cdot \left(16 + 10 y\right) - 4 y = 20$

Then develop the brackets:

$- 48 - 30 y - 4 y = 20$

Solve for $y$:

$- 34 y = 68$

$y = - 2$

Then replace $y$ in 2) by what you found:

$x - 10 \cdot \left(- 2\right) = 16$

Solve for $x$:

$x = - 4$

Finished!