Writing and Graphing Equations in Standard Form

1.10.3 Writing and Graphing Equations in Standard Form

1.10.3 • Writing and Graphing Equations in Standard Form

Activity

For questions 1 – 3, use the following situation:

Andre’s coin jar contains 85 cents. There are no quarters or pennies in the jar, so the jar has all nickels, all dimes, or some of each.

1.

Write an equation that relates the number of nickels, nn, the number of dimes, dd, and the amount of money, in cents, in the coin jar.

Compare your answer:

5n+10d=855n+10d=85, where n is the number of nickels and d is the number of dimes.

2.

Graph your equation on the coordinate plane. Be sure to label the axes. Use the graphing tool or technology outside the course. Graph the equation that represents this scenario using the Desmos tool below.

Compare your answer:

A graph with number of nickels on the x-axis and number of dimes on the y-axis. A red line slopes downward from left to right, passing through the points (0,8.5) and (17,0).
3.

How many nickels are in the jar if there are no dimes?

Compare your answer:

17 nickels

4.

How many dimes are in the jar if there are no nickels?

Compare your answer:

It's not possible for the coin jar to have no nickels if the amount of money in it is 85 cents.

Are you ready for more?

Extending Your Thinking

What are all the different ways the coin jar could have 85 cents if it could also contain quarters?

Compare your answer:

Listed as nickels, dimes, quarters: (17,0,0), (15,1,0), (13,2,0), (12,0,1), (11,3,0), (10,1,1), (9,4,0), (8,2,1), (7,0,2), (7,5,0), (6,3,1), (5,1,2),(5,6,0), (4,4,1), (3,2,2), (2,0,3), (3,7,0), (2,5,1), (1,3,2), (0,1,3), (1,8,0), (0,6,1)

Self Check

A coin jar contains 75 cents and only contains quarters and nickels. Write an equation that relates the number of quarters, q , the number of nickels, n , and the amount of money, in cents, in the coin jar.
  1. 0.05 n − 0.25 q = 0.75
  2. 25 q − 5 n = 75
  3. 25 q + 5 n = 0.75
  4. 25 q + 5 n = 75

Additional Resources

Standard Form of Linear Equation

A linear equation is in standard form when it is written Ax+By=CAx+By=C.

Most people prefer to have AA, BB, and CC be integers and A≥0A≥0 when writing a linear equation in standard form, although it is not strictly necessary.

  • Linear equations have infinitely many solutions.
  • For every number that is substituted for xx there is a corresponding yy value.
  • This pair of values is a solution to the linear equation and is represented by the ordered pair (x,y)(x,y).
  • When we substitute these values of xx and yy into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.

Solution of a Linear Equation in Two Variables

An ordered pair (x,y)(x,y) is a solution of the linear equation Ax+By=CAx+By=C, if the equation is a true statement when the xx– and yy–values of the ordered pair are substituted into the equation.

Linear equations have infinitely many solutions. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions.

A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to that equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions!

When a linear equation in standard form is solved for yy, it is then easy to pick out the slope of the line and where the graph of the line crosses the yy–axis.

Example

Solve 9x−3y=159x−3y=15 for y

Step 1 - Move the xx term to isolate the yy term.

Subtract 9x9x from both sides.

9x−9x−3y=15−9x9x−9x−3y=15−9x

Step 2 - Simplify.

−3y=15−9x−3y=15−9x

Step 3 - Divide by the coefficient of yy.

Divide both sides by −3−3

−3y−3=15−9x−3−3y−3=15−9x−3

Step 4 - Simplify.

y=3x−5y=3x−5

In the new equation, y=3x−5y=3x−5, the slope is 33 and the graph of the line would cross the yy–axis at -5.

Try it

Understanding Standard Form

Solve 6x+2y=146x+2y=14 for yy.

Compare your answer:

Step 1 - Move the xx term to isolate the yy term.

Subtract 6x6x from both sides.

6x−6x+2y=14−6x6x−6x+2y=14−6x

Step 2 - Simplify.

2y=−6x+142y=−6x+14

Step 3 - Divide by the coefficient of yy.

Divide both sides by 2.

2y2=−6x+1422y2=−6x+142

Step 4 - Simplify.

y=−3x+7y=−3x+7