From e2e6c48041d37999b5adf77500b7aec2e57604e6 Mon Sep 17 00:00:00 2001 From: zleyyij Date: Fri, 2 Feb 2024 09:18:16 -0700 Subject: [PATCH] vault backup: 2024-02-02 09:18:16 --- education/math/Systems of Equations.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/education/math/Systems of Equations.md b/education/math/Systems of Equations.md index 657ab73..45243a7 100644 --- a/education/math/Systems of Equations.md +++ b/education/math/Systems of Equations.md @@ -4,3 +4,25 @@ | Inconsistent | Parallel lines, no solution | | Independent | The lines only cross at one point. | | Dependant | The lines are identical, and there are infinitely many solutions. Both equations represent the same line when plotted. | +# Solving +## Graphing +Graph the two equations, and look for points where they intersect + +## Substitution +Solve for a variable in the equation, then substitute it into another equation. +EG, given the below systems: +$$ y = x+3 $$ +$$ x = 3y + 2 $$ +You can substitute values, like this: +$$ x = 3(x+3) + 2 $$ +Then once you have $x$, you can plug it back into the first equation to solve for y. + +## Elimination +To solve a system of equations by elimination, you add the two equations together and see if something cancels out. +A simple example might look like this: +$$ 2x + 3y = 23 $$ +$$ 4x -3y = -13 $$ +You "add" the two equations together, getting: +$$ 6x + 0y = 10 $$ +You now know that $6x = 10$. +If you don't have two values that evenly cancel out, like $3$ and $4$, you can find the least common multiple and multiply the entire equation so that those two are equal. In this case, you'd multiply one equation by 4, and one equation by 3. \ No newline at end of file