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

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Answer 1 · 2016-02-20T23:00:38

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

$2 y + 6 = 3 y + 10$ $2y + 6 = 3y + 10$

$2 y - 3 y = 10 - 6$ $2y - 3y = 10 - 6$

$- y = 4$ $-y = 4$

$y = - 4$ $y = -4$

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

$2 (- 4) + 6 = x$ $2(-4) + 6 = x$

$- 8 + 6 = x$ $-8 + 6 = x$

$- 2 = x$ $-2 = x$

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

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

Always solve for the easiest variable in the equation. For example, in $6 x + 3 y = - 9$ $6x + 3y = -9$ , 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.
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 $4 x - 2 y = - 3$ $4x - 2y = -3$ , you must only replace y:

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

You would then distribute and then solve.

Practice exercises:

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

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

Good luck!

How do you solve the system of linear equations 2y+6=x2y+6=x and 3y+10=x3y+10=x?