The Slope of the Line y=mx+b

The Slope of the Line y=mx+b exercise

• Recall how to write equations in slope-intercept form.

  Hints

  Isolate the variable $y$ to put the linear relationship into slope-intercept form.

  Subtract the $x$ term from both sides of the equation.

  Divide every term by the coefficient of $y$ to cancel it out.

  Solution

  All linear relationships can be written in slope-intercept form, $y=mx+b$, by isolating the variable $y$.

  1. For the koala example, the linear relationship is $3y=500x$.

  • Divide both sides by $3$.
  • $y=\frac{500}{3}x$

  2. For the pig example, the linear relationship is $120x-4y=20$.

  • Subtract $120x$ from both sides of the equation.
  • $-4y=20-120x$
  • Divide all terms by $-4$.
  • $y=-5+30x$.
  • Rearrange the terms on the right.
  • $y=30x-5$

  3. For the honey badger example, the linear relationship is $3x+5y=10$.

  • Subtract $3x$ from both sides of the equation.
  • $5y=10-3x$.
  • Divide all terms by 5.
  • $y=2-\frac{3}{5}x$
  • Rearrange the terms on the right.
  • $y=-\frac{3}{5}x+2$

• Identify the slope and the $y$-intercept of each equation.

  Hints

  Slope-intercept form is $y=mx+b$.

  The slope is $m$ and the $y$-intercept is $b$.

  To get $y$ alone, subtract the $x$ term if necessary, then divide each term by the coefficient of $y$.

  Solution

  Slope-intercept form is $y=mx+b$ where $m$ is the slope and $b$ is the $y$-intercept. To find $m$ and $b$, put all of the equations in slope-intercept form $y=mx+b$.

  • $y=2x+1$ is in slope-intercept form. Therefore, the slope is $2$ and the $y$-intercept is $1$.

  • $y=\frac{8}{5}x$ is in slope-intercept form. Therefore, the slope is $\frac{8}{5}$ and the $y$-intercept is $0$ since there is no $b$ in the equation.

  • $y=-\frac{1}{2}x-3$ is in slope-intercept form. Therefore the slope is $-\frac{1}{2}$ and the $y$-intercept is -3.

  • $4x+y=1$ is not in slope-intercept form. Subtracting $4x$ from both sides of the equation yields $y=1-4x$. Then, rearrange the terms on the right so the equation is $y=-4x+1$. Now it looks like $y=mx+b$ which means the slope is $-4$ and the $y$-intercept is $1$.

  • $-2y=3x-6$ is not in slope-intercept form. To get $y$ alone, divide every term by $-2$: $y=-\frac{3}{2}+3$. The equation is now in the form $y=mx+b$ which means the slope is $-\frac{3}{2}$ and the $y$-intercept is $3$.

• Match the equations with their slope-intercept form.

  Hints

  Isolate the variable $y$ by subtracting the $x$ term from both sides.

  Rearrange the terms or flip the equation to make it look like $y=mx+b$.

  Divide every term by the coefficient of $y$.

  Solution

  1. $-2y+3x=8$

  • Subtract $3x$ from both sides
  • $-2y=8-3x$
  • Divide every term by $-2$
  • $y=-4+\frac{3}{2}x$
  • Rearrange the terms to look like $y=mx+b$
  • $y=\frac{3}{2}x-4$

  2. $12x+3y=6$

  • Subtract $12x$ from both sides
  • $3y=6-12x$
  • Divide every term by $3$
  • $y=2-4x$
  • Rearrange the terms on the right
  • $y=-4x+2$

  3. $\frac{1}{3}y=2x$

  • Multiply both sides by $3$
  • $y=6x$

  4. $5y=20x+30$

  • Divide every term by 5
  • $y=4x+6$

  5. $9x+8=3y$

  • Flip the equation
  • $3y=9x+8$
  • Divide every term by $3$
  • $y=3x+\frac{8}{3}$

  6. $4x=6y$

  • Flip the equation
  • $6y=4x$
  • Divide both sides by $6$ and simplify
  • $y=\frac{2}{3}$

• Graph the equation $-y-3x+4=0$.

  Hints

  The graph of a line is $y=mx+b$, where $m$ is the slope and $b$ is the $y$-intercept.

  To graph the $y$-intercept, plug $x=0$ into the equation.

  The slope, $m$, is the change in $y$ divided by the change in $x$.

  Solution

  The line $-y-3x+4=0$ can be rewritten in slope-intercept form by solving for $y$.

  $~$

  To solve for $y$:

  1. Add $3x$ to both sides: $-y+4=3x$.

  2. Subtract $4$ from both sides: $-y=3x-4$.

  3. Divide both sides by $-1$: $y=-3x+4$.

  $~$

  To identify the slope and $y$-intercept:

  • The slope is $m=-3$.
  • The $y$-intercept is $b=4$.

  $~$

  To identify the point at $x=1$:

  • Start at the $y$-intercept, $(0,4)$.
  • As the slope is $m=-3$, if we subtract $3$ from the $y$-coordinate and add $1$ to the $x$-coordinate of a point on our line, we get another point on our line.
  • Doing this with our $y$-intercept, we get $(0+1,4-3)=(1,1)$, which is the point on our line at $x=1$.
  • Use the slope $m=-3$ to find the point at $x=1$.

  $~$

  To graph:

  1. Plot the $y$-intercept $(0,4)$.

  2. Plot the point $(1,1)$.

  3. Draw a line connecting these two points.

• Determine if the equation is written in slope-intercept form or not.

  Hints

  Slope-intercept form is $y=mx+b$.

  Slope-intercept form can also look like $y=b+mx$, $mx+b=y$ or $b+mx=y$.

  If there is no $y$-intercept $b$, slope-intercept form will look like $y=mx$.

  Solution

  Slope-intercept form is $y=mx+b$. However, the terms can be rearranged and still be in slope-intercept form. The following equations are $4$ different ways to represent slope-intercept form.

  • $y=mx+b$: the original slope-intercept form
  • $y=b+mx$: the terms on the right side are switched
  • $mx+b=y$: the equation is flipped so that y is on the right
  • $b+mx=y$: the equation is flipped with y on the right, and the $mx$ and $b$ terms are switched

  If there is no $y$-intercept, the equation will look like $y=mx$ or $mx=y$.

  The following equations are in slope-intercept form because they match one of the equations listed above.

  • $y=5x-3$
  • $y=4+2x$
  • $y=-\frac{2}{5}x$
  • $y=-4x$
  • $1-3x=y$
  • $-x+1=y$

  The following equations are not in slope-intercept form because they do not match any of the slope-intercept form equations above.

  • $x=-\frac{1}{2}y+1$
  • $\frac{3}{2}x+y=2$
  • $2y=x+1$
  • $x=-\frac{1}{2}y+1$
  • $-y=2x-3$
  • $7y=2x$

• Determine the line.

  Hints

  Slope-intercept form is $y=mx+b$, where $m$ is the slope and $b$ is the $y$-intercept.

  If $x=0$ in the point $(x,y)$, then the $y$-coordinate is the $y$-intercept $b$.

  If $x\neq 0$ in the point $(x,y)$, substitute the slope $m$ and the point $(x,y)$ into $y=mx+b$, and solve for $b$.

  Solution

  Slope-intercept form, $y=mx+b$, consists of a point $(x,y)$, a slope $m$, and a $y$-intercept $b$. Given a point and a slope you can uniquely determine a line.

  1. $(0,-4)$, $m=\frac{1}{2}$

  • Since $x=0$, we know that the $y$-intercept is $-4$, $b=-4$
  • Now we can substitute $m$ and $b$ into $y=mx+b$
  • $y=\frac{1}{2}x-4$

  2. $(0,\frac{35}{2})$, $m=\frac{3}{5}$

  • Since $x=0$, we know that the $y$-intercept is $\frac{35}{2}$, $b=\frac{35}{2}$
  • Now we can substitute $m$ and $b$ into $y=mx+b$
  • $y=\frac{3}{5}x+\frac{35}{2}$

  3. $(1,2)$, $m=2$

  • Since $x\neq 0$, we substitute the values of $x$, $y$, and $m$ into $y=mx+b$, to solve for $b$.
  • $2=2(1)+b$
  • $2=2+b$
  • $0=b$
  • Now that we know $b=0$, we can substitute $b$ and $m$ into $y=mx+b$
  • $y=2x+0$
  • $y=2x$

  4. $(-1,-3)$, $m=-2$

  • Since $x\neq 0$, we substitute the values of $x$, $y$, and $m$ into $y=mx+b$, to solve for $b$.
  • $-3=-2(-1)+b$
  • $-3=2+b$
  • $-5=b$
  • Now that we know $b=-5$, we can substitute $b$ and $m$ into $y=mx+b$
  • $y=-2x-5$

  5. $(0,-6)$, $m=-\frac{17}{3}$

  • Since $x=0$, we know that the $y$-intercept is $-6$, $b=-6$
  • Now we can substitute $m$ and $b$ into $y=mx+b$
  • $y=-\frac{17}{3}x-6$

  6. $(4,4)$, $m=6$

  • Since $x\neq 0$, we substitute the values of $x$, $y$, and $m$ into $y=mx+b$, to solve for $b$.
  • $4=6(4)+b$
  • $4=24+b$
  • $-20=b$
  • Now that we know $b=-20$, we can substitute $b$ and $m$ into $y=mx+b$
  • $y=6x-20$

  7. $(-4,3)$, $m=\frac{1}{4}$

  • Since $x\neq 0$, we substitute the values of $x$, $y$, and $m$ into $y=mx+b$, to solve for $b$.
  • $3=\frac{1}{4}(-4)+b$
  • $3= -1+b$
  • $4=b$
  • Now that we know $b=4$, we can substitute $b$ and $m$ into $y=mx+b$
  • $y=\frac{1}{4}x+4$