4.05 Standard form
There are a number of ways of stating an equation for a straight line. Previously, we saw slope-intercept form and we will see others in future lessons. Now, we'll discover the value of writing equations in standard form.
The standard form of a linear equation is
$Ax+By=C$Ax+By=C
where $A$A, $B$B, and $C$C are all integers and the value of $A$A is positive, that is, $A>0$A>0.
When we are given an equation in standard form, we can either graph using intercepts or rearrange it to one of our other forms.
Graphing using intercepts
The standard form of a line is great for identifying both the $x$x and $y$y intercepts.
For example, the line $2x+3y=6$2x+3y=6
The $x$x intercept happens when the $y$y value is $0$0.
| $2x+3y$2x+3y | $=$= | $6$6 |
| $2x+3\left(0\right)$2x+3(0) | $=$= | $6$6 |
| $2x$2x | $=$= | $6$6 |
| $x$x | $=$= | $3$3 |
The $y$y intercept happens when the $x$x value is $0$0.
| $2x+3y$2x+3y | $=$= | $6$6 |
| $2\left(0\right)+3y$2(0)+3y | $=$= | $6$6 |
| $3y$3y | $=$= | $6$6 |
| $y$y | $=$= | $2$2 |
Rearranging to slope-intercept form
If we don't want to graph using intercepts, we can also rearrange the equation to slope-intercept form. Let's look at rearranging between the two forms.
Worked example
Express the equation $4x+6y=12$4x+6y=12 in slope-intercept form.
Think: We need to solve for $y$y to get the equation to the form $y=mx+b$y=mx+b.
Do: To solve for $y$y, we need to reverse the operations to get $y$y by itself.
| $4x+6y$4x+6y | $=$= | $12$12 | Start with the given equation |
| $6y$6y | $=$= | $-4x+12$−4x+12 | Undo addition by subtracting $4x$4x from both sides |
| $y$y | $=$= | $\frac{-4x+12}{6}$−4x+126 | Undo multiplication by dividing by $6$6 on both sides |
| $y$y | $=$= | $\frac{-2}{3}x+2$−23x+2 | Simplify |
Reflect: $y$y is isolated, so the linear equation $4x+6y=12$4x+6y=12 is $y=\frac{-2}{3}x+2$y=−23x+2 in slope-intercept form.
Rearranging to standard form
As a mathematical convention, we are often asked to give our answer in standard form. Let's look at rearranging to standard form.
Worked example
Express the equation $y=\frac{-3}{4}x-5$y=−34x−5 in standard form.
Think: We need get to the form $Ax+By=C$Ax+By=C, where $A$A, $B$B and $C$C are integers and $A>0$A>0.
Do: We need to ensure there are no non-integer coefficients and that $A$A is positive.
| $y$y | $=$= | $\frac{-3}{4}x-5$−34x−5 | Start with the given equation |
| $4y$4y | $=$= | $-3x-20$−3x−20 | Clear the fraction by multiplying by $4$4 |
| $3x+4y$3x+4y | $=$= | $-20$−20 | Add $3x$3x to both sides to get the x and y terms to the same side |
Reflect: We now have something of the form $Ax+By=C$Ax+By=C, where $A$A, $B$B and $C$C are integers and$A>0$A>0, so the linear equation $y=\frac{-3}{4}x-5$y=−34x−5 is $3x+4y=-20$3x+4y=−20 in standard form.
Practice questions
question 1
Express the following equations in standard form.
$y=6x-5$y=6x−5
$y=\frac{6x}{5}-6$y=6x5−6
question 2
Consider the line given by the equation: $5x-3y=-15$5x−3y=−15
Solve for $x$x-value of the $x$x-intercept of the line.
Solve for $y$y-value of the $y$y-intercept of the line.
Hence, graph the equation of the line.
Loading Graph...
question 3
A line has slope $\frac{5}{7}$57 and passes through the point $\left(-3,-4\right)$(−3,−4).
By substituting into the equation $y=mx+b$y=mx+b, find the value of $b$b for this line.
Write the equation of the line in standard form.