# How do you solve the system 2x - 5y = 5 and 2x + 2y = 26?

Dec 10, 2017

We subtract the left side of equation 2 from the left side of equation 1, and the right side of equation 2 from the right side of equation 1. See explanation.

#### Explanation:

General Method: For a more general solution for this type of problem, consider:

${a}_{1} x + {b}_{1} y = {c}_{1}$ is ${E}_{1}$
${a}_{2} x + {b}_{2} y = {c}_{2}$ is ${E}_{2}$

To perform the operation here, we must combine the two rows in some manner that only gets rid of one variable. Supposing we wish to remove the x variable by subtracting some multiple of the second equation from the first, we would multiply the left and right hand sides of equation 2 by ${a}_{1} / {a}_{2}$, and subtract the left hand side of the result from the left hand side of the first equation, and the right hand side of the result from the right hand side of the first equation:

$\to {a}_{1} / {a}_{2} {E}_{2} = {a}_{1} / {a}_{2} \left({a}_{2} x\right) + {a}_{1} / {a}_{2} \left({b}_{2} y\right) = {a}_{1} / {a}_{2} \left({c}_{2}\right) \to {a}_{1} \left(x\right) + {a}_{1} / {a}_{2} \left({b}_{2} y\right) = {a}_{1} / {a}_{2} {c}_{2}$

Subtracting:

${E}_{1} - {a}_{1} / {a}_{2} {e}_{2} = \left({a}_{1} - {a}_{1}\right) x + \left({b}_{1} - {a}_{1} / {a}_{2} {b}_{2}\right) y = {c}_{1} - {a}_{1} / {a}_{2} {c}_{2}$

$\to \left({b}_{1} - {a}_{1} / {a}_{2} {b}_{2}\right) y = {c}_{1} - {a}_{1} / {a}_{2} {c}_{2}$

Divide by the coefficient of the y term on both sides:

$\to y = \frac{{c}_{1} - {a}_{1} / {a}_{2} {c}_{2}}{{b}_{1} - {a}_{1} / {a}_{2} {b}_{2}}$

Having found y, we would plug it into either equation to find x.

Solution to this specific problem:Because our coefficient on the x terms for both equations is the same (i.e. 2), the first thing we can do is simply subtract the second equation from the first. We do this as follows:

${E}_{1} - {E}_{2} \to \left(2 - 2\right) x + \left(- 5 - 2\right) y = \left(5 - 26\right)$

Giving us...

$\to - 7 y = - 21 \to - 7 \frac{y}{-} 7 = - \frac{21}{-} 7 \to y = 3$

Substitute this back into our first or second equation (we will use first here):

$2 x - 5 \left(3\right) = 5 \to 2 x - 15 = 5 \to 2 x = 20 \to x = 10$

Thus our solution is $\left(10 , 3\right)$

Plug this into each equation to check your work.

${E}_{1} : 2 \left(10\right) - 5 \left(3\right) = 5 \to 20 - 15 = 5$ is accurate..

${E}_{2} : 2 \left(10\right) + 2 \left(3\right) = 26 \to 20 + 6 = 26$ is also accurate.

Thus our solution of $\left(10 , 3\right)$ is correct.