How do you solve the system of linear equations 2y+6=x and 3y+10=x?

Feb 20, 2016

Since x is already isolated in both equations we can easily solve for y by substitution.

Explanation:

$2 y + 6 = 3 y + 10$

$2 y - 3 y = 10 - 6$

$- y = 4$

$y = - 4$

Now solve for x by substituting -4 for y in either one of the equations.

$2 \left(- 4\right) + 6 = x$

$- 8 + 6 = x$

$- 2 = x$

The solution set is $\left\{- 2 , - 4\right\}$.

Here are a few of the most fundamental rules for solving systems of equations by substitution.

1. Always solve for the easiest variable in the equation. For example, in $6 x + 3 y = - 9$, it would be easiest to solve for y because you don't end up with fractions, which can become long and tricky to work with.

2. Always only replace the value of x or y in the equation. Don't get rid of any coefficients x or y may have. Example:

If you want to substitute 2x + 3 = y into $4 x - 2 y = - 3$, you must only replace y:

#4x - 2(2x + 3) = -3

You would then distribute and then solve.

Practice exercises:

1. Solve the following linear systems of equations by substitution.

a). $x + 3 y = - 3 , 2 x - 2 y = 10$

b). $2 x + 3 y = 10 , - 3 x + 4 y = 36$

Good luck!