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

$2x+y=11\,$ $-4x+3y=13\,$ Multiply 1st row by 2 add to 2nd row

$4x+2y=22\,$ $-4x+3y=13\,$ Yields

$5y=35\,$ => $y=7$ Multiply 1st row by -3 add to 2nd row

$-6x-3y=-33\,$ $-4x+3y=13\,$ Yields

$-10x=-20\,$ => $x=2$ Substitution

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

$2x+y=11\,$ $-4x+3y=13\,$ You can switch around some terms in the first to get this:

$y=-2x+11\,$ Then you can substitute that into the bottom one so that it looks like this:

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

$2x+y=11\,$ $-4x+3y=13\,$ Solve for y

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

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