Algebra I: Writing Linear Equations

Case A: you are given b

Slope $-4$ and $y$-intercept $6\text{.}$

• Nothing to solve: $m=-4$ and $b=6$ go straight in.

• $y=-4x+6\text{.}$

The $y$-intercept is the point $(0, 6)\text{,}$ the one point where substituting is free.

Case B: you are given another point

Slope $\tfrac{1}{2}$ through $(4, 7)\text{.}$

• Substitute: $7=\tfrac{1}{2}(4)+b=2+b\text{.}$

• So $b=5$ and $y=\tfrac{1}{2}x+5\text{.}$

Any point works the same way; $b$ is whatever makes that point fit.

Case A: a point with a negative

Through $(-1, 6)$ with slope $-2\text{.}$

• $y-6=-2(x-(-1))\text{.}$

• Subtracting a negative is adding: $y-6=-2(x+1)\text{.}$

The signs in the form are minus signs, so a negative coordinate shows up as a plus.

Case B: two points

Through $(1, 2)$ and $(3, 8)\text{.}$

• Slope first: $m=\dfrac{8-2}{3-1}=\dfrac{6}{3}=3\text{.}$

• Use either point: $y-2=3(x-1)\text{,}$ which becomes $y=3x-1\text{.}$

With $(3, 8)$ instead: $y-8=3(x-3)$ also becomes $y=3x-1\text{.}$ Same line.

Case A: standard to slope-intercept

$4x-2y=10\text{.}$

• Subtract $4x\text{:}$ $-2y=-4x+10\text{.}$

• Divide every term by $-2\text{:}$ $y=2x-5\text{.}$

Dividing by a negative flips the sign of every term.

Case B: slope-intercept to standard

$y=\tfrac{2}{3}x-1\text{.}$

• Multiply every term by $3$ to clear the fraction: $3y=2x-3\text{.}$

• Move the $x$-term to the left: $-2x+3y=-3\text{,}$ or $2x-3y=3$ after multiplying by $-1\text{.}$

Standard form usually has whole numbers and a positive $A\text{.}$

Case A: two points or a table

From $(-2, 7)$ to $(4, -5)\text{:}$ $m=\dfrac{-5-7}{4-(-2)}=\dfrac{-12}{6}=-2\text{.}$

• A table works the same way: if $x$ is $2, 4, 6$ and $y$ is $1, 4, 7\text{,}$ then $y$ goes up $3$ while $x$ goes up $2\text{,}$ so $m=\tfrac{3}{2}\text{.}$

Change in $y$ over change in $x\text{,}$ every time.

Case B: a graph

Pick two points where the line crosses grid corners, such as $(0, 1)$ and $(3, 3)\text{.}$

• Run: $3$ to the right. Rise: $2$ up.

• $m=\tfrac{2}{3}\text{.}$

Falling to the right means a negative rise, so a negative slope.

Case A: a table

$x$ is $1, 2, 3$ and $y$ is $7, 10, 13\text{.}$

• $y$ goes up $3$ each time $x$ goes up $1\text{,}$ so $m=3\text{.}$

• There is no $x=0\text{,}$ so walk back one step: $7-3=4\text{.}$ So $b=4$ and $y=3x+4\text{.}$

A table hides $b$ in the $x=0$ column, even when that column isn't shown.

Case B: a story

A $24$ cm candle burns down $1.5$ cm every hour.

• The rate is the slope, and burning down makes it negative: $m=-1.5\text{.}$

• The start, at hour $0\text{,}$ is $b=24\text{.}$ So $h=-1.5t+24\text{.}$

Words like per, each and every mark the rate.

Case A: parallel

Through $(-1, 3)\text{,}$ parallel to $y=-3x+7\text{.}$

• Same slope: $m=-3\text{.}$

• $3=-3(-1)+b=3+b\text{,}$ so $b=0$ and $y=-3x\text{.}$

A $b$ of $0$ is a real answer: this line passes through the origin.

Case B: perpendicular

Through $(3, 1)\text{,}$ perpendicular to $y=\tfrac{3}{4}x-2\text{.}$

• Flip and change the sign: $m=-\tfrac{4}{3}\text{.}$

• $1=-\tfrac{4}{3}(3)+b=-4+b\text{,}$ so $b=5$ and $y=-\tfrac{4}{3}x+5\text{.}$

Check: $\tfrac{3}{4}\cdot\left(-\tfrac{4}{3}\right)=-1\text{.}$

Case A: horizontal

Through $(1, -4)$ and $(5, -4)\text{.}$

• Slope: $\dfrac{-4-(-4)}{5-1}=\dfrac{0}{4}=0\text{.}$

• The equation is $y=-4\text{.}$

Zero rise means slope $0\text{.}$

Case B: vertical

Through $(3, 1)$ and $(3, 6)\text{.}$

• Slope: $\dfrac{6-1}{3-3}=\dfrac{5}{0}\text{,}$ which is undefined.

• The equation is $x=3\text{.}$

Zero run means undefined slope.