The Graph of a Linear Equation in Two Variables

The Graph of a Linear Equation in Two Variables exercise

• Complete the table for the equation $y+\frac{1}{2}x=5$.

  Hints

  Plug in each value for $x$ into the equation $y+\frac{1}{2}x=5$ and solve for $y$.

  Plugging in $x=4$ into $y+\frac{1}{2}x=5$, we get $y+\frac{1}{2} \cdot {4}=5$, which reduces to $y+2=5$.

  Subtracting $2$ from both sides gives us $y=3$.

  Solution

  To complete the table using the given $x$-values, plug in each of these $x$-values in the equation beside and isolate $y$:

  • $x=0$ leads to $y+\frac12(0)=5$, which simplifies to $y=5$.
  • $x=3$ leads to $y+\frac12(3)=5$, which simplifies to $y+1.5=5$. Subtracting $1.5$ from both sides gives us $y=3.5$.
  • $x=5$ leads to $y+\frac12(5)=5$, which simplifies to $y+2.5=5$. Here we subtract $2.5$ from both sides to get $y=2.5$.
  • $x=8$ leads to $y+\frac12(8)=5$, which simplifies to $y+4=5$. Subtracting $4$ from both sides results in $y=1$.
  • $x=10$ leads to $y+\frac12(10)=5,$ which simplifies to $y+5=5$. We have to subtract $5$ from both sides to get $y=0$.

• Determine which statements about linear equations are true.

  Hints

  When writing down a linear equation, such as $2y^1-3x^1=5$, the exponents of the variables do not need to be written down, as they are always $1$: $2y^1-3x^1 = 2y-3x$.

  Here you see a parabola, which is the graph of a quadratic function.

  Solution

  A linear equation is a first-degree equation.

  What does this mean? It means that all variables in a linear equation are raised to the power of $1$.

  The graph of a linear equation is always a straight line.

• Find the table which belongs to each linear equation.

  Hints

  Check each given $x$ by plugging it into the linear equation.

  If any ordered pair $(x,y)$ from a table is not a solution to a specific equation, then that equation cannot represented by that table. That means you can move on an try another equation.

  Perhaps it's easier if you rewrite each equation as we do with the following equation:

  $\begin{array}{rcl} y-2x & = & ~4\\ \color{#669900}{+2x} & &\color{#669900}{2x}\\ y & = & ~2x+4 \end{array}$

  Solution

  Let's check each table and the corresponding equation from left to the right:

  The first table

  $y-2x=4$? We plug in the given $x$-values into this equation to get:

  • for $x=2$, we have $y-2(2)=4$. Adding $4$ leads to $y=8$. $~~~~$✓
  • for $x=3$, we get $y-2(3)=4$, and thus $y=10$. $~~~~$✓

  Similarly, we can check $x=5$, $x=6$, and $x=8$. $~~~~$✓

  $y-2x=4$ is the linear equation which is represented by this table.

  Likewise, you get the following matches:

  • The second table represents $y+\frac12x=0$.
  • The third table represents $y-x=4$.
  • The fourth table represents $y+2x=3$.
  • The fifth table represents $y+\frac12x=3$.

• Graph the linear equation $y-2x=2$.

  Hints

  You can create a table by taking the $x$-coordinate of each possible point on the graph, and plugging it into the equation to get the corresponding $y$-coordinate. If the pair $(x,y)$ matches one of the possible points, then that point lies on the desired line.

  Three points belong to the given equation.

  Solution

  To draw a line corresponding to a given linear equation, you proceed as follows:

  • Create a table: take some values for $x$ and plug those values into the given equation. Solve this equation for $y$. You get a bunch of ordered pairs $(x,y)$, which are entries in your table.
  • Highlight those points in a coordinate plane.
  • Connect those points to get the line which is represented by the given equation.

• Identify which table belongs to which graph.

  Hints

  For each table, every row gives you an ordered pair $(x,y)$.

  There is one decreasing line; its equation is $y+x=0$.

  The other equations are:

  • $y-x=0$
  • $y-1.5x=0$
  • $y-2x=-1$

  Solution

  Each row of a table gives you an ordered pair $(x,y)$. We can plot these points on the coordinate plane as follows:

  • Draw a line parallel to the $y$-axis that passes through the $x$-coordinate of $(x,0)$.
  • Draw a line parallel to the $x$-axis that passes through the $y$-coordinate of $(0,y)$.
  • The intersection of those lines is the point $(x,y)$.

  Once you have plotted all of the points from the table, if you notice that they all lie on a line, you can connect all of the points to form this line.

  We get the following lines:

  • The first table belongs to the green line. The corresponding equation is given by $y-x=0$.
  • The second table belongs to the blue line and its equation is $y-1.5x=0$.
  • The third table belongs to the red line and is represented by the linear equation $y+x=0$.
  • The fourth table belongs to the yellow line. The corresponding equation is given by $y-2x=-1$.

• Determine which linear equation belongs to each graph.

  Hints

  Write each equation in the form $y=mx+b$, where $m$ is the slope and $b$ the $y$-intercept.

  Check the given points by plugging them in the given equations.

  Solution

  Here you see a detailed solution for the line pictured here.

  You can either check the given points, $(2,2)$ and $(4,3)$, or determine the corresponding linear equation explicitly.

  If you choose to determine the corresponding linear equation, then we first note that it will be of the form $y = ax + b$, where $b$ is the $y$-intercept and $a$ is the slope.

  • We can see from the picture that the $y$-intercept is $b=1$.
  • So we still have to determine $a$. For this, put a point on the line into the equation $y = ax + 1$ For example, $(2,2)$. You can also decide to choose the other point. The equation will be the same. You can test that yourself!

  This gives us the following equation: $2 = a(2) + 1$ which we solve using opposite operations:

  $\begin{array}{rcl} 2 & = & 2a + ~1\\ \color{#669900}{-1} & &\color{#669900}{-1}\\ 1 & = & 2a\\ \color{#669900}{\div2} & & \color{#669900}{\div2}\\ ~\frac12 & = & a \end{array}$

  That's it! The desired equation is thus given by $y = \frac12x + 1$, which is the same as $ y - \frac12x = 1 $.

  The equation corresponding to the leftmost graph is given by $y-2x=-1$.

  The equation corresponding to the second graph from the right is given by $y+\frac12x=3.5$.

  The equation corresponding to the rightmost graph is given by $y-\frac32x=-0.5$.