New Zealand

Level 6 - NCEA Level 1

Lesson

So far we've had a look at what simultaneous equations are and at some of the ways to solve them. Here is a quick recap.

As with all algebraic expressions, simultaneous equations can also be expressed as graphs on a number plane. In coordinate geometry, we know that each graph represents ALL the possible solutions of a related equation. In other words, if a point is on a graph, it must solve its equation. In simultaneous equations, we mostly deal with linear equations, which can be represented as straight line graphs. Our aim is then to find a solution that solves BOTH equations, and graphically this means finding the point of intersection of the two straight lines.

Let's have a look at an example, where we want to find the solution to the simultaneous equations $y=5x$`y`=5`x` and $y=x+2$`y`=`x`+2. So then we would plot the two equations as graphs. Remember there are two ways to visualise linear equations as graphs: either through finding its intercepts or finding its gradient-intercept form. Here I have drawn the $y=5x$`y`=5`x` line as red and the $y=x+2$`y`=`x`+2 line as green:

We can then see that there is only one intersection point and it is $(0.5,2.5)$(0.5,2.5). Therefore the solution that solves the two equations must be when $x=0.5$`x`=0.5 and $y=2.5$`y`=2.5.

Simultaneous equations are easy to solve if we know how to graph them, but that takes a long time and sometimes we need a faster way. One efficient way to solve them is called the substitution method. As the simultaneous equations we usually work with involve $2$2 variables and $2$2 equations, this method works by solving one variable first through 'substituting' one equation into the other.

To solve something like $3y+4=x$3`y`+4=`x` and $2+x-y=0$2+`x`−`y`=0 we'd first have to pick one equation to transform so that either $x$`x` or $y$`y` is the subject. The second equation looks more easy to manipulate, and transforming it gives us $y=2+x$`y`=2+`x`. Knowing the exact value of $y$`y` means we can combine the two equations by substituting $y=2+x$`y`=2+`x` into $3y+4=x$3`y`+4=`x`. This gives us:

$3\left(2+x\right)+4$3(2+x)+4 |
$=$= | $x$x |

$6+3x+4$6+3x+4 |
$=$= | $x$x |

$3x-x$3x−x |
$=$= | $-4-6$−4−6 |

$2x$2x |
$=$= | $-10$−10 |

$x$x |
$=$= | $-5$−5 |

Now that we know the value of $x$`x` we can easily solve for $y$`y` by using the equation we already have that has $y$`y` as the subject: $y=2+x$`y`=2+`x`. Therefore $y=2+\left(-5\right)$`y`=2+(−5) which equals $-3$−3. So our answer is $(-5,-3)$(−5,−3).

The elimination method works by adding or subtracting equations (yes you can do that!) to eliminate one variable so that we only have one variable to solve at a time.

Now we know how to eliminate variables from a system of simultaneous equations by adding or subtracting.

For example I can add $2x-y=1$2`x`−`y`=1 and $5x+y=2$5`x`+`y`=2 to get $7x=3$7`x`=3, which doesn't have any $y$`y`'s.

OR

I can subtract $3y-x=2$3`y`−`x`=2 from $3y-2x=0$3`y`−2`x`=0 to get $-x=-2$−`x`=−2, which also doesn't have any $y$`y`'s.

We can see that we should use addition when the coefficients of the variable we want to eliminate are equal and opposite in sign, and subtraction when they're equal and the same sign. But what happens when they don't have the same coefficients at all?

When we don't have the same value coefficients for the variables we want to eliminate, we can multiply or divide the whole equation by a constant until it gets to the coefficient that we want.

For example, we have $x+3y=5$`x`+3`y`=5 & $2y+2x=1$2`y`+2`x`=1. Say we want to eliminate the $x$`x`'s first. However, the coefficient of $x$`x` is $1$1 in one equation and $2$2 in the other. The idea then is to multiply the first equation by $2$2 so that we have an $x$`x` coefficient of $2$2 for both equations. Then we can apply the same logic as before and add or subtract the equations. Let's see this in action:

Multiplying the first equation by $2$2:

$2\left(x+3y\right)$2(x+3y) |
$=$= | $2\times5$2×5 |

$2x+6y$2x+6y |
$=$= | $10$10 |

Now let's find the difference of this new equation and Equation 2:

$2x+6y-\left(2y+2x\right)$2x+6y−(2y+2x) |
$=$= | $10-1$10−1 |

$2x+6y-2y-2x$2x+6y−2y−2x |
$=$= | $9$9 |

$4y$4y |
$=$= | $9$9 |

$y$y |
$=$= | $\frac{9}{4}$94 |

Now that we have our value for $y$`y` it's a simple case of substituting it back in any of the equations to get our value for $x$`x`:

$x+3y$x+3y |
$=$= | $5$5 |

$x+\frac{3\times9}{4}$x+3×94 |
$=$= | $5$5 |

$x+\frac{27}{4}$x+274 |
$=$= | $5$5 |

$x$x |
$=$= | $5-\frac{27}{4}$5−274 |

$x$x |
$=$= | $\frac{-7}{4}$−74 |

Consider the following two equations: $y=2x+2$`y`=2`x`+2 & $y=-2x+2$`y`=−2`x`+2

a) What are the $x$`x`-intercepts and $y$`y`-intercepts of $y=2x+2$`y`=2`x`+2?

b) What are the $x$`x`-intercepts and $y$`y`-intercepts of $y=-2x+2$`y`=−2`x`+2?

c) Graph the two equations on the same number plane.

d) State the values of $x$`x` and $y$`y` that satisfy both equations.

Use the substitution method to solve for $p$`p` and $q$`q`.

a) First solve for $q$`q`.

b) Now solve for $p$`p`.

Use the elimination method to solve for $x$`x` & $y$`y`:

Equation 1 | $4x-5y=6$4x−5y=6 |
- $(1)$(1) |

Equation 2 | $2x-y=3$2x−y=3 |
- $(2)$(2) |

a) Solve for $x$`x`

Think: What's the easiest way to to make the coefficients of the $y$`y` terms be the same value, of equal or opposite sign?

Do:

The $y$`y` coefficients of the two equations are $-5$−5 and $-1$−1, so let's multiply Equation 2 by $5$5 to make it $-5$−5 as well.

$(2)\times5$(2)×5: | |||

$5\left(2x-y\right)$5(2x−y) |
$=$= | $5\times3$5×3 | |

$10x-5y$10x−5y |
$=$= | $15$15 | - $(3)$(3) |

Let's name this Equation 3. Now that the coefficients of $y$`y` in Equations 1 & 3 are equal in value and sign, let's subtract one from the other. Let's try Equation 3 take away Equation 1:

$(3)-(1)$(3)−(1): | ||

$10x-5y-\left(4x-5y\right)$10x−5y−(4x−5y) |
$=$= | $15-6$15−6 |

$10x-5y-4x+5y$10x−5y−4x+5y |
$=$= | $9$9 |

$6x$6x |
$=$= | $9$9 |

$x$x |
$=$= | $\frac{3}{2}$32 |

b) Solve for $y$`y`

Think: Which of the $3$3 equations we have would be easiest to derive $y$`y` from, if we have $x$`x`?

Do:

Equation 2 seems the most simple so let's substitute our $x$`x` value in there:

$x$x→$(2)$(2): |
||

$2\times\frac{3}{2}-y$2×32−y |
$=$= | $3$3 |

$3-y$3−y |
$=$= | $3$3 |

$3$3 | $=$= | $y+3$y+3 |

$y$y |
$=$= | $0$0 |

Check:

Subbing in our $x$`x` and $y$`y` values into our original equations:

$4x-5y$4x−5y |
$=$= | $6$6 |

$\frac{4\times3}{2}-5\times0$4×32−5×0 | $=$= | $6$6 |

$2\times3-0$2×3−0 | $=$= | $6$6 |

$6$6 | $=$= | $6$6 |

$2x-y$2x−y |
$=$= | $3$3 |

$\frac{2\times3}{2}-0$2×32−0 | $=$= | $3$3 |

$3$3 | $=$= | $3$3 |

Use the elimination method by subtraction to solve for $x$`x` and then $y$`y`.

Equation 1 | $2x-5y=1$2x−5y=1 |

Equation 2 | $-3x-5y=-39$−3x−5y=−39 |

First solve for $x$

`x`Now solve for $y$

`y`

Based on the diagram given, find the value of $x$`x` and $y$`y`.

Use the fact that $AB=CD$

`A``B`=`C``D`to set up Equation 1.Make sure to simplify your equation by grouping any like terms.

Use the fact that $AD=BC$

`A``D`=`B``C`to set up Equation 2.Make sure to simplify your equation by grouping any like terms.

First solve for $x$

`x`.Equation 1 $64x-3y=372$64 `x`−3`y`=372Equation 2 $8x-15y=-12$8 `x`−15`y`=−12Now solve for $y$

`y`.Equation 1 $64x-3y=372$64 `x`−3`y`=372Equation 2 $8x-15y=-12$8 `x`−15`y`=−12Find the length of the rectangle.

Find the width of the rectangle.