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Solving formulas 2
Example
`y=mx+b` for m
`y=mx+b` | Solving for m, focus on addition first |
`ycolor(red)(-b)=mx+bcolor(red)(-b)` | Subtract `b` from both sides |
`y-b=mx` | `m` is multipied by `x`. |
`(y-b)/color(red)(m)=(mx)/color(red)(m) ` | Divide both sides by x |
`(y-b)/x=m` | solution |
It is important to note that we know we are done with the problem when the variable we are solving for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation.
Example
`h = (2m)/n` for `m`
`h = (2m)/n` | To clear the fraction |
`color(red)((n))h = color(red)((n))(2m)/n` | Multiply each term by `n` |
`nh=2m` | Reduce `n` with denominators |
`(nh)/color(red)(2)=(2m)/color(red)(2) ` | Divide both sides by `2` |
`(nh)/2=m ` | Solution |
Example
`a/b + c/b = e` for `a`
`a/b + c/b = e` | To clear the fraction we use `LCD = b` |
`a/bcolor(red)((b)) + c/bcolor(red)((b)) = ecolor(red)((b))` | Multiply each term by `b` |
`a+c = eb` | Reduce `b` with denominators |
`a+c color(red)(-c) = eb color(red)(-c)` | Subtract `c` from both sides |
`a=eb−c ` | Solution |
Example
`a= A /(2-b)` for `b`
`a= A /(2-b)` | Use `(2-b)` as a group |
`a color(red)((2-b))= A /(2-b)color(red)((2-b))` | Multiply each term by (2 − b) |
`(2−b)a=A` | reduce `(2 − b)` with denominator |
`2a−ab=A` | Distribute through parenthesis |
`2a color(red)(-2a)−ab=A color(red)(-2a)` | Subtract 2a from both sides |
`−ab=A−2a` | The `b` is multipied by `−a` |
`(−ab)/color(red)((-a))=(A−2a)/color(red)((-a))` | Divide both sides by `− a` |
`b = (A − 2a)/(−a)` | Solution |
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