How do solve the following linear system?:  -7x+y=-19 , -2x+3y=-1 ?

Feb 27, 2017

$\left(x , y\right) = \left(\frac{56}{19} , \frac{31}{19}\right)$

Explanation:

$\left\{\begin{matrix}- 7 x + y = - 19 \\ - 2 x + 3 y = - 1\end{matrix}\right.$

Notice that the single variable $y$ is easily isolated in the first equation.

Taking the first equation $- 7 x + y = - 19$, we can rewrite this by adding $7 x$ to both sides to see that

$\textcolor{b l u e}{y = 7 x - 19}$

We now have an expression equal to $y$ completely in terms of $x$. We can then take the equation we haven't used yet and replace $y$ with $7 x - 19$, since for this system we know these are equivalent expressions.

$- 2 x + 3 \textcolor{b l u e}{y} = - 1 \text{ "=>" } - 2 x + 3 \textcolor{b l u e}{\left(7 x - 19\right)} = - 1$

We can now solve this equation, since it's entirely in terms of $x$. Distributing the $3$ into the parentheses gives

$- 2 x + \left(21 x - 57\right) = - 1$

Combining the $x$ terms then adding $57$ to both sides gives

$19 x - 57 = - 1$

$19 x = 56$

Then

color(red)(x=56/19

Now we can plug this into either equation to find the value of $y$:

$- 2 \textcolor{red}{x} + 3 y = - 1$

$- 2 \textcolor{red}{\left(\frac{56}{19}\right)} + 3 y = - 1$

$- \frac{112}{19} + 3 y = - 1$

Multiplying everything by $19$ yields

$- 112 + 57 y = - 19$

So

$57 y = 93$

$y = \frac{93}{57}$

Both of these are divisible by $3$:

$y = \frac{31 \times 3}{19 \times 3}$

$\textcolor{g r e e n}{y = \frac{31}{19}}$

So the solution is the point $\left(\textcolor{red}{x} , \textcolor{g r e e n}{y}\right) = \left(\textcolor{red}{\frac{56}{19}} , \textcolor{g r e e n}{\frac{31}{19}}\right)$.

$\text{ }$

The graphs of the the lines $- 7 x + y = - 19$ and $- 2 x + 3 y = - 1$ should intersect at the point $\left(\frac{56}{19} , \frac{31}{19}\right) \approx \left(2.95 , 1.63\right)$:

graph{(-7x+y+19)(-2x+3y+1)=0 [-10.61, 17.87, -4.8, 9.44]}

They do!