# Fundamental Mathematics/Arithmetic/Linear equation/Roots of System of Linear Equations

## Roots of System of Linear Equations

One was mentioned above, but there are other ways to solve a system of linear equations without graphing.

### Variable Elimination

${\displaystyle 2x+y=11\,}$
${\displaystyle -4x+3y=13\,}$

Multiply 1st row by 2 add to 2nd row

${\displaystyle 4x+2y=22\,}$
${\displaystyle -4x+3y=13\,}$

Yields

${\displaystyle 5y=35\,}$ => ${\displaystyle y=7}$

Multiply 1st row by -3 add to 2nd row

${\displaystyle -6x-3y=-33\,}$
${\displaystyle -4x+3y=13\,}$

Yields

${\displaystyle -10x=-20\,}$ => ${\displaystyle x=2}$

### Substitution

If you get a system of equations that looks like this:

${\displaystyle 2x+y=11\,}$
${\displaystyle -4x+3y=13\,}$

You can switch around some terms in the first to get this:

${\displaystyle y=-2x+11\,}$

Then you can substitute that into the bottom one so that it looks like this:

${\displaystyle -4x+3(-2x+11)=13\,}$
${\displaystyle -4x-6x+33=13\,}$
${\displaystyle -10x+33=13\,}$
${\displaystyle -10x=-20\,}$
${\displaystyle x=2\,}$

Then, you can substitute 2 into an x from either equation and solve for y. It's usually easier to substitute it in the one that had the single y. In this case, after substituting 2 for x, you would find that y = 7.

### Determinant

If you get a system of equations that looks like this:

${\displaystyle 2x+y=11\,}$
${\displaystyle -4x+3y=13\,}$

Solve for y

${\displaystyle x+{\frac {1}{2}}y={\frac {11}{2}}\,}$
${\displaystyle x+{\frac {3}{-4}}y={\frac {13}{-4}}\,}$
${\displaystyle y({\frac {1}{2}}-{\frac {3}{4}})={\frac {11}{2}}-{\frac {13}{-4}}}$
${\displaystyle y={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{4}}}}}$

Solve for x

${\displaystyle 2x+y=11\,}$
${\displaystyle {\frac {-4}{3}}x+y={\frac {13}{3}}\,}$
${\displaystyle x(2-{\frac {-4}{3}})=11-{\frac {13}{3}}}$
${\displaystyle x={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{4}}}}}$