There are two ways to handle simultaneous equations.
This is the addition (or subtraction) method. Simultaneous Equations involve two equations with two unknowns. The substitution method involves:
• Rearrange one equation so that one unknown is on one side
• In the other equation, replace the unknown from the first equation
• Solve this second equation (with one unknown)
• Replace the solved value in the first equation
• Solve the first equation
• Check the values are correct
Note that it is not important which equation is used to start.
If one equation is a multiple of another, then there will be no solution (i.e. they are parallel lines).
Solve the simultaneous equations `4x-y=6` and `5x+2y=1`
Rearrange the first equation to get it in terms of `y`
| `4x` | `-` | `y` | `=` | `6` | |||
| Add `y` to both sides: | `4x` | `=` | `6` | `+` | `y` | ||
| Subtract 6 from both sides: | `4x` | `-` | `6` | `=` | `y` |
Replace `y` in the second equation `4x-6`
| Second equation: | `5x` | `+` | `2y` | `=` | `1` | ||
| substitute for `y`: | `5x` | `+` | `2(4x-6)` | `=` | `1` | ||
| expand brackets: | `5x` | `+` | `8x` | `-` | `12` | `=` | `1` |
| add x terms: | `13x` | `-` | `12` | `=` | `1` | ||
| add 12 to both sides: | `13x` | `=` | `13` | ||||
| divide both sides by 13 | `x` | `=` | `1` |
Replace `x` in the first equation with the found value of `x`
| First equation: | `4x` | `-` | `y` | `=` | `6` | ||
| substitute for `x`: | `4(1)` | `-` | `y` | `=` | `6` | ||
| `4` | `-` | `y` | `=` | `6` | |||
| add `y` to both sides: | `4` | `=` | `6` | `+` | `y` | ||
| subtract 6 from both sides: | `-2` | `=` | `y` |
Finally, check by putting both values in the other equation
| Check using 2nd equation: | `5(1)+2(-2)=1`✔ |
Answer: `x=1, y=-2`
Solve the simultaneous equations `4x+3y=14` and `6x+2y=11`.
Rearrange the first equation to be in terms of `y`
| 2nd equation to get `y` | `6x` | + | `2y` | = | 11 | ||
| subtract 6x from both sides: | `2y` | = | 11 | - | `6x` | ||
| divide all terms by 2 | `y` | = | `frac(11)(2)` | - | `3x` |
Solve the equation for `x`
| First equation: | `4x` | + | `3y` | = | 14 | ||
| Replace the value `y`: | `4x` | + | `3(frac(11)(2)-3x)` | = | 14 | ||
| Multiply out the brackets: | `4x` | `+` | `frac(33)(2)` | `-` | `9x` | `=` | `14` |
| subtract the fraction from both sides: | `4x` | `` | `` | `-` | `9x` | `=` | `-frac(5)(2)` |
| add the `x`-values: | `-5x` | `` | `` | `` | `` | `=` | `-frac(5)(2)` |
| divide both sides by 5: | `-x` | `` | `` | `` | `` | `=` | `-frac(1)(2)` |
| multiply both sides by -1: | `x` | `` | `` | `` | `` | `=` | `frac(1)(2)` |
Substitute the found value of `x`
| Second equation | `6x` | `+` | `2y` | `=` | `11` | ||
| Substitute known value of `x` | `6(frac(1)(2))` | `+` | `2y` | `=` | `11` | ||
| `3` | `+` | `2y` | `=` | `11` | |||
| Subtract 3 from both sides | `` | `` | `2y` | `=` | `8` | ||
| Divide both sides by 2 | `` | `` | `y` | `=` | `4` |
Check the answer
| Check using 1st equation | 4(`frac(1)(2)`)+3(4)=14` ✔ |
Answer: `x=frac(1)(2)` and `y=4`