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Graphs of linear equations are useful for describing relationships between variables that change at a constant rate. They often outperform words or equations. Keeping in mind the definition of a linear equation in one variable, it will now be extended to two variables. This lesson will also cover the most basic method of graphing linear equations in two variables.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Consider the following linear equations.
The solutions of Equations (I), (II), and (III) are $x=1,$ $x=2,$ and $x=3,$ respectively. The points corresponding to each equation will be plotted on one coordinate plane. First, the points must be identified. Their $x-$coordinates will be the solutions and their $y-$coordinates will be the left-hand sides of the original related equations.

Now the points can be plotted on a coordinate plane.

$3=2x+15=2x+17=2x+1 (I)(II)(III) $

The right-hand sides of these equations are all the same. Now, each equation will be solved using the Properties of Equality.
$3=2x+15=2x+17=2x+1 $

Solve for $x$

$(I), (II), (III):$ $LHS−1=RHS−1$

$2=2x4=2x6=2x $

$(I), (II), (III):$ $LHS/2=RHS/2$

$1=x2=x3=x $

$(I), (II), (III):$ Rearrange equation

$x=1x=2x=3 $

Equation | Solution | Point |
---|---|---|

$3=2x+1$ | $x=1$ | $(1,3)$ |

$5=2x+1$ | $x=2$ | $(2,5)$ |

$7=2x+1$ | $x=3$ | $(3,7)$ |

What can be noted about the points? What is the equation of the line that passes through them? Why?

A table of values is a chart that helps to organize and visualize information. It is frequently used to show the relation between two variables.

$x$ | $y$ |
---|---|

$0$ | $0$ |

$1$ | $3$ |

$2$ | $6$ |

$3$ | $9$ |

$4$ | $12$ |

$5$ | $15$ |

$2$corresponds to the $y-$value

$6.$This is usually represented with the notation $(2,6).$

The relation between the variables of an equation can be shown by making a table of values. This is helpful when graphing any equation, not only linear ones. The following linear equation will be drawn as an example.
*expand_more*
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*expand_more*

$6x−3y=-12 $

There are four steps to making a table of values for an equation.
1

Isolate One Variable

If in the given equation none of the variables are isolated on one side, it is convenient to do so before making the table of values. This will simplify the rest of the process. A variable can be isolated using the Properties of Equality. If the variables presented in the equation are $x$ and $y,$ usually the latter one will be isolated.
Often the non-isolated variable is called the input and the isolated variable is called the output.

$6x−3y=-12$

SubEqn

$LHS−6x=RHS−6x$

$-3y=-6x−12$

DivEqn

$LHS/(-3)=RHS/(-3)$

$y=-3-6x−12 $

Simplify right-hand side

WriteSumFrac

Write as a sum of fractions

$y=-3-6x +-3-12 $

DivNegNeg

$-b-a =ba $

$y=36x +312 $

CalcQuot

Calculate quotient

$y=2x+4$

2

Choose the Values for the Non-Isolated Variable(s)

A table of values will usually have as many columns as there are variables __plus one__. In this case, the first column is for the values of the input. The second column is for substituting values from the first column into the rewritten equation. The third column shows the values of the output corresponding to its input.

$x$ | $2x+4$ | $y=2x+4$ |
---|

Next, the values of the non-isolated input variable $x$ should be chosen. It can be done arbitrarily. However, the values should always belong to the domain of the function represented by the given equation. Recall that the domain of a linear function are all real numbers. In this case $-2,$ $0,$ $0.5,$ and $1$ will be used.

$x$ | $2x+4$ | $y=2x+4$ |
---|---|---|

$-2$ | ||

$0$ | ||

$0.5$ | ||

$1$ |

3

Substitute the Values Into the Rewritten Equation

In this step the input values will be substituted into the rewritten equation where the output variable is isolated on the left-hand side.

$x$ | $2x+4$ | $y=2x+4$ |
---|---|---|

$-2$ | $2(-2)+4$ | |

$0$ | $2(0)+4$ | |

$0.5$ | $2(0.5)+4$ | |

$1$ | $2(1)+4$ |

4

Calculate the Values of the Isolated Variable

The last step is to simplify the expressions in the second column and write the result in the the last column. These are the outputs corresponding to the input from each row.

$x$ | $2x+4$ | $y=2x+4$ |
---|---|---|

$-2$ | $2(-2)+4$ | $0$ |

$0$ | $2(0)+4$ | $4$ |

$0.5$ | $2(0.5)+4$ | $5$ |

$1$ | $2(1)+4$ | $6$ |

If the expressions in the second column are too complicated to calculate mentally, extra columns where the partial results are written can be added as needed. However, the last column should always be the outputs.

$x$ | $2x+4$ | Simplify | $y=2x+4$ |
---|---|---|---|

$-2$ | $2(-2)+4$ | $-4+4$ | $0$ |

$0$ | $2(0)+4$ | $0+4$ | $4$ |

$0.5$ | $2(0.5)+4$ | $1+4$ | $5$ |

$1$ | $2(1)+4$ | $2+4$ | $6$ |

The table of values can be reduced to a table with only two columns — one with inputs and one with outputs. Note that for the output column we write only the variable.

$6x−3y=-12$ | |
---|---|

$x$ | $y$ |

$-2$ | $0$ |

$0$ | $4$ |

$0.5$ | $5$ |

$1$ | $6$ |

Zain is considering a new phone plan. Currently they spend $$15$ on their phone each month. The new plan consists of a $$9$ flat rate for which they get unlimited texts, $300$ minutes, and $6GB$ of data. For any minute above the $300$ minutes included in the plan, Zain will have to pay an additional $$0.01.$

In the previous months Zain used $350,$ $410,$ and even $550$ minutes. However, in the past they never exceeded $600$ minutes. Make a table of values and help Zain decide if they should switch to the new plan.{"type":"choice","form":{"alts":["Yes","No"],"noSort":false},"formTextBefore":"\"Should Zain switch to the new plan?\"","formTextAfter":"","answer":0}

Let $y$ represent the total cost of the new phone plan. Let $x$ represent the number of minutes used. Write an equation for $y$ when $x$ is *greater than* $300.$

Let $y$ represent the total cost of the new phone plan expressed in dollars. Let $x$ represent the number of minutes used. If $x$ is *less than or equal to* $300,$ $y$ is equal to the flat rate of $$9.$
*greater than* $300,$ the cost of the minutes exceeding the $300$ given minutes must be added to the $flat$ $rate.$ There are $x−300$ minutes that have to be payed for. Each of these minutes costs $$0.01.$
Now, the table of values for $x$ and $y$ will be formed. Let $x$ be $350,$ $410,$ and $550.$ These are the numbers of minutes Zain used in the previous months. Since they never exceeded $600$ minutes, this will be the highest considered value of $x.$ Note that all considered values of $x$ are *greater than* $300,$ so the expression for $y$ is $0.01x+6.$

$x≤300⇒y=9 $

If $x$ is $x>300⇒y=9+(x−300)0.01 $

The above equation for $y$ can be simplified.
$y=9+(x−300)0.01$

$y=0.01x+6$

$x$ | $y=0.01x+6$ | $y$ |
---|---|---|

$350$ | $y=0.01(350)+6$ | $9.5$ |

$410$ | $y=0.01(410)+6$ | $10.1$ |

$550$ | $y=0.01(550)+6$ | $11.5$ |

$600$ | $y=0.01(600)+6$ | $12$ |

For the full picture, the values of $x=200$ and $x=300$ will also be added to the table. Remember that when $x≤300,$ the value of $y$ is always equal to $9.$

Number of Minutes | Total Cost |
---|---|

$200$ | $$9$ |

$300$ | $$9$ |

$350$ | $$9.5$ |

$410$ | $$10.1$ |

$550$ | $$11.5$ |

$600$ | $$12$ |

Currently Zain spends $$15$ on their phone plan. It can be noted that all of the prices in the above table are *less than* $$15.$ Assuming Zain does not change their talking habits and $6GB$ is enough data for them, they __should__ switch to the new plan.

A linear equation is an equation with at least one linear term and any number of constants. No other types of terms may be included. Linear equations in one variable have the following form, where $a$ and $b$ are real numbers and $a =0.$

$ax+b=0orax=b $

Linear equations in two variables have the form below, where $a,$ $b,$ and $c$ are real numbers and $a =0$ and $b =0.$

$ax+by+c=0orax+by=c $

A line is a one-dimensional object of infinite length with no width or thickness that never bends or turns. Its graphical representation is a straight line with arrowheads on either end, indicating that it continues indefinitely in both directions.

Through any two different points there is exactly one line. A line can be named using any two points on it. The above line could be named $AB,$ $BA,$ or line $ℓ.$ When two or more points lie on the same line, they are said to be collinear.The graph of a linear equation in two variables is the set of all its solutions plotted on the same coordinate plane, forming a line. Since a line is infinite, this means that a linear equation in two variables has infinitely many solutions.

Consider the given equation in two variables. Determine whether it is a linear equation or not.

Ramsha works in a bakery. She earns $$10$ per hour. Since it is December, she will get a bonus of $$50$ this month.

The following linear equation describes Ramsha's pay.$y=10x+50 $

Here $y$ represents Ramsha's pay in dollars, and $x$ represents the number of hours she works. Using a table of values, graph the given linear equation.
Choose *at least* two $x-$values for the table of values. Since $x$ represents the number of hours worked, it must be *greater than* $0.$

The given linear equation will be graphed using a table of values.
*at least* two different values of $x$ need to be evaluated in the table of values. Also, $x$ must be *greater than* $0$ as it represents the number of hours worked.

$y=10x+50 $

The graph of a linear equation in two variables is a line. There is exactly one line that passes through any two different points. Therefore, $x$ | $y=10x+50$ | $y$ |
---|---|---|

$35$ | $y=10(35)+50$ | $400$ |

$95$ | $y=10(95)+50$ | $1000$ |

$125$ | $y=10(125)+50$ | $1300$ |

Each pair of $x$ and $y$ corresponds to an $(x,y)$ coordinate pair. Now $(35,400),$ $(95,1000),$ and $(125,1300)$ will be plotted on the same coordinate plane.

Finally, the three points can be connected using a straightedge.

Since in the given case $x$ cannot be negative, the graph includes only one arrowhead and is actually a ray.

Maya loves reading. She is buying books at a garage sale.

The following linear equation describes the relationship between the amount of money $y$ in dollars that Maya will have left after buying $x$ books.$3x+2y=46 $

Make a table of values for this equation and then graph it. What is the maximum number of books that Maya could afford?
**Graph:**

**Maximum Number of Books:** $15$

Choose *at least* two $x-$values for the table of values.

First, the given equation will be graphed. Then the obtained graph will be used to determine the maximum number of books that Maya can purchase.

$3x+2y=46$

Solve for $y$

SubEqn

$LHS−3x=RHS−3x$

$2y=-3x+46$

DivEqn

$LHS/2=RHS/2$

$y=2-3x+46 $

WriteSumFrac

Write as a sum of fractions

$y=2-3x +246 $

CalcQuot

Calculate quotient

$y=-1.5x+23$

$x$ | $y=-1.5x+23$ | $y$ |
---|---|---|

$0$ | $y=-1.5(0)+23$ | $23$ |

$2$ | $y=-1.5(2)+23$ | $20$ |

$4$ | $y=-1.5(4)+23$ | $17$ |

$10$ | $y=-1.5(10)+23$ | $8$ |

Note that in order to simplify the graphing process, the chosen $x-$values are all even numbers. Each pair of $x$ and $y$ corresponds to a coordinate pair $(x,y).$ Now $(0,23),$ $(2,20),$ $(4,17),$ and $(10,8)$ will be plotted on the same coordinate plane.

Finally, the four points can be connected using a straightedge.

Since the amount $y$ of money left and the number $x$ of books purchased cannot be negative, the graph is bounded only to the first quadrant. This also means that only part of the line is actually considered, which makes it a segment.

Note that the number of books has to be a whole number, as Maya cannot buy part

of a book. Also, she cannot spend more money than she has, which means that the amount of money left has to be non-negative. A point which meets these conditions, the maximum value of the $x-$coordinate, can be identified on the graph.

The $x-$coordinate of the point corresponds to the maximum number of books that Maya can buy. As seen on the graph, its value is $15.$ Therefore, Maya can purchase *at most* $15$ books.

Equations in one variable can be solved by using the properties of equality. Consider the following equation in one variable.
*at least* two different values of $x$ have to be evaluated in the table.

This value of $x$ can be substituted into the original equation and simplified. If both sides of the equation are equal after simplifying, this value is a solution to the equation.
Both sides of the original equation equal $5$ for $x=2.$ Therefore, $x=2$ is a solution to the equation.

$2x+1=5x−5 $

This type of equation can also be solved by graphing the linear equations in two variables corresponding to the left- and right-hand sides of the given equation.
$2x+1=5x−5⇒y=2x+1y=5x−5 (I)(II) $

Equations (I) and (II) will be graphed on one coordinate plane using a table of values. First, the table of values for Equation (I) will be made. Remember that $x$ | $y=2x+1$ | $y$ |
---|---|---|

$0$ | $y=2(0)+1$ | $1$ |

$1$ | $y=2(1)+1$ | $3$ |

The line passing through $(0,1)$ and $(1,3)$ is the graph of $y=2x+1.$ Now the table for Equation (II) will be constructed.

$x$ | $y=5x−5$ | $y$ |
---|---|---|

$0$ | $y=5(0)−5$ | $-5$ |

$1$ | $y=5(1)−5$ | $0$ |

The line passing through $(0,-5)$ and $(1,0)$ is the graph of $y=5x−5.$

Finally, the point of intersection of $y=2x+1$ and $y=5x−5$ can be identified.

The lines intersect at $(2,5).$ This means that both $2x+1$ and $5x−5$ equal $5$ when $x=2.$

Expression | $x=2$ | Simplify |
---|---|---|

$2x+1$ | $2(2)+1$ | $5$ |

$5x−5$ | $5(2)−5$ | $5$ |