The Graph of a Linear Equation in Two Variables Is a Line

The Graph of a Linear Equation in Two Variables Is a Line exercise

• Describe how to graph the line corresponding to the equation $y=\frac13x +12$.

  Hints

  A general linear equation has the form, $y=x+b$, where $m$ is the slope and $b$ the $y$-intercept.

  The $y$-intercept lies on the $y$-axis, while the $x$-intercept lies on the $x$-axis.

  Here we see the solution of an equation similar to the equation in this problem.

  Solution

  There is an easy way to graph the line corresponding to this equation: determine the intercepts of both coordinate axes and draw the line which goes through both points.

  Keep in mind that you only need two points to draw a line uniquely.

  The $y$-intercept has the $x$-coordinate $x=0$.

  We plug this into the equation, $y=\frac13x +12$, to get:

  $y=\frac13(0)+12=0+12=12$.

  So the point where the line crosses the $y$-axis is $(0,12)$.

  Generally, the $y$-intercept is the term without $x$ in the linear equation: $y=mx+b$, namely $b$.

  The x-intercept has the $y$-coordinate $y=0$.

  We plug this into $y=\frac13x +12$ to get $0=\frac13x+12$.

  We solve this equation using opposite operations:

  $\begin{array}{rcl} 0 & = & ~\frac13x+12\\ \color{#669900}{-12} & &\color{#669900}{-12}\\ -12 & = & ~\frac13x\\ \color{#669900}{\times 3} & & \color{#669900}{\times 3}\\ -36 & = & ~x \end{array}$

  So the point where the line crosses the $x$-axis is $(-36,0)$.

• Decide which statements about linear equations are true.

  Hints

  The $y$-intercept is the $y$-coordinate of the point where the line crosses the $y$-axis.

  Just look at the $y$-axis and the numbers on the $x$-axis:

  • On the left of the $y$-axis we have $x=-1$.
  • On the right of the $y$-axis we have $x=1$.

  Here you see an example of a linear equation. Plugging $x=0$ into the equation leads to $y=4(0)+3=0+3=3$.

  This is a quadratic equation.

  Solution

  Here you see a linear equation in a generalized form:

  • $m$ is the slope.
  • $b$ is the $y$-intercept.

  Both variables, $x$ as well as $y$, are raised to the power of $1$.

  So, let's have a look at the coordinate-intercepts:

  To get the $y$-intercept, just plug $x=0$ into the equation to get:

  $y=m(0)+b=0+b=b$.

  To get the $x$-intercept, you have to solve the equation $0=mx+b$.

  Subtract $b$ to get $-b=mx$, then divide both sides by $m$ to get the $x$-intercept $x=-\frac bm$.

• Identify the $x$- and $y$-intercepts for each line.

  Hints

  The $x$-coordinate of the $y$-intercept is $x=0$.

  The $x$-intercept has the $y$-coordinate, $y=0$.

  Solution

  The $y$-intercept of a line is where it intersects the $y$-axis, and the $x$-intercept of a line is where it intersects the $x$-axis. With this, we can read off the $x$- and $y$-intercepts of each line:

  $~$

  The first line

  • $y$-intercept: $(0,4)$
  • $x$-intercept: $(-4,0)$

  $~$

  The second line

  • $y$-intercept: $(0,2)$
  • $x$-intercept: $(5,0)$

  $~$

  The third line

  • $y$-intercept: $(0,6)$
  • $x$-intercept: $(-2,0)$

  $~$

  The fourth line

  • $y$-intercept: $(0,4)$
  • $x$-intercept: $(2,0)$

• Find the $x$- and $y$-intercepts for the given equations.

  Hints

  The $y$-intercept of $y=mx+b$ is $(0,b)$.

  To solve this equation:

  • Subtract $b$ on both sides to get $-b=mx$.
  • Divide both sides by $m$ to get $x=-\frac bm$.

  Solution

  In general, this equation has the $y$-intercept $(0,b)$, and the $x$-intercept $\left(-\frac bm,0\right)$.

  So for the given equations we have the following $x$- and $y$-intercepts:

  $\mathbf{y=\frac12x-4}$

  • For $x=0$, we get $y=-4$ and so $(0,-4)$ is the $y$-intercept.
  • $x=-\frac{-4}{0.5}=8$, which gives us $(8,0)$ as the $x$-intercept.

  $\mathbf{y=2x+4}$

  • $(0,4)$ is the $y$-intercept.
  • $x=-\frac42=-2$ leads to $(-2,0)$ as the $x$-intercept.

  $\mathbf{y=-3x-12}$

  • $(0,-12)$ is the $y$-intercept.
  • $x=-\frac{-12}{-3}=-4$ gives us the $x$-intercept $(-4,0)$.

  $\mathbf{y=-\frac12x-2}$

  • $(0,-2)$ is the $y$-intercept.
  • $x=-\frac{-2}{-0.5}=-4$ gives us the $x$-intercept $(-4,0)$.

• Plot the ordered pairs on a coordinate plane.

  Hints

  Keep in mind that the first coordinate of an ordered pair is the $x$-coordinate, and the second one is the $y$-coordinate.

  To draw a point proceed as follows:

  • Draw a line parallel to the $y$-axis, which intersects the given $x$-coordinate.
  • Draw a line parallel to the $x$-axis, which intersects the given $y$-coordinate.
  • The intersection of these two lines is the desired point.

  Solution

  Here you see the correct ordered pairs. To plot them on the coordinate plane, first remember:

  • The left-most coordinate is the $x$-coordinate.
  • The right-most coordinate is the $y$-coordinate.

  Draw any ordered pair or point as follows:

  • Draw a line parallel to the $y$-axis, which intersects the given $x$-coordinate.
  • Draw a line parallel to the $x$-axis, which intersects the given $y$-coordinate.
  • The intersection of these two lines is the desired point.

  So you get the points we want:

  • $(0,6)$ the green one on the $y$-axis
  • $(3,4)$ the blue one
  • $(6,2)$ the violet one
  • $(9,0)$ the yellow one on the $x$-axis

  Can you recognize that all given points are lying on one line? The equation for this line is $y=-\frac23x+6$.

• Graph the line for the equation $f(x)=\frac{20}{3}x+20$.

  Hints

  First check the $y$-intercept. This is the constant term of the equation.

  You have to solve the equation, $0=mx+b$, to get the $x$-intercept. This leads to $x=-\frac bm$.

  Solution

  The linear equation is given by $y=\frac{20}3x+20$.

  We can see the $y$-intercept from the equation: $b=20$. This is the point $(0,20)$.

  To get the $x$-intercept, we have to solve the following equation:

  $\begin{array}{rcl} 0 & = & ~\frac{20}3x+20\\ \color{#669900}{-20} & &\color{#669900}{-20}\\ -20 & = & ~\frac{20}3x\\ \color{#669900}{\times \frac3{20}} & & \color{#669900}{\times \frac3{20}}\\ -3 & = & ~x \end{array}$

  So the $x$-intercept is the point $(-3,0)$.

  Drawing the line which goes through these two points gives us the line corresponding to the equation, $y=\frac{20}3x+20$.