Precalculus by Richard Wright

Previous Lesson Table of Contents Next Lesson

Are you not my student and
has this helped you?

This book is available
to download as an epub.


Blessed are those who mourn, for they will be comforted. Matthew‬ ‭5‬:‭4‬ ‭NIV‬‬

8-02 Two-Variable Linear Systems

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.1

Acoustic guitar
Figure 1: Acoustic guitar. credit (pixabay/9883074)

An instrument company makes guitars and is starting production on a new model. It has a one-time cost of $10,050 to set up the factory production line and materials cost $150 per guitar. The company is going to sell the the guitars for $300 each. How many guitars do they need to sell to break even, where the costs equal the revenue?

To solve this problem, create an equation for costs and one for revenue. This will be a system of equations to be solved. In this lesson, the method will be elimination. Unlike substitution, elimination only works for linear equations.

Linear Systems of Equations

Linear systems of equations in two-variables produce three possible types of solutions. The lines may intersect in one point. The equations may graph the same line, so all the points on the line are solutions. Or, the line may be parallel and never intersect, so there is no solution. If there is at least one solution, the system is consistent. If there is only one solution, it is independent, but many solutions is dependent because the y-value of the solution depends on the x-value. If there is no solution, the system is inconsistent.

Type of Solutions of Systems
1 solution
1 Solution: Consistent and Independent
many solutions
Infinitely Many Solutions: Consistent and Dependent
no solution
No Solution: Inconsistent

Elimination

Solve Linear Equations by Elimination
  1. Write the equations in columns (ax + by = c).
  2. Obtain coefficients of one variable that differ only in sign by multiplying the equations by constants.
  3. Add the equations and solve the resulting equation. (A variable will be eliminated.)
  4. Back-substitute the answer into either original equation and solve.
  5. Check your solution.

If all the variables are eliminated and the result is

Example 1: Solve a Linear System

Solve \(\left\{\begin{align} 4x + y &= -3 \\ x - 3y &= 9 \end{align}\right.\) by elimination.

Solution

The system is already written in columns. If the first equation were multiplied by 3, then the coefficients of y would be 3 and −3.

$$ \left\{\begin{align} 3\left(4x + y\right) &= 3\left(-3\right) \\ x - 3y &= 9 \end{align}\right. $$

$$ \left\{\begin{align} 12x + 3y &= -9 \\ x - 3y &= 9 \end{align}\right. $$

Add the equations

$$ \begin{align} 12x + 3y &= -9 \\ + \quad \underline{x - 3y} &= \underline{9} \\ 13x \qquad &= 0 \end{align} $$

$$ x = 0 $$

Back-substitute this into one of the original equations.

$$ 4x + y = -3 $$

$$ 4(0) + y = -3 $$

$$ y = -3 $$

The solution seems to be (0, −3). Check this by plugging it into both of the original equations.

$$ \left\{\begin{align} 4x + y &= -3 \\ x - 3y &= 9 \end{align}\right. $$

$$ \left\{\begin{align} 4(0) + (-3) &= -3 \\ (0) - 3(-3) &= 9 \end{align}\right. $$

$$ \left\{\begin{align} -3 &= -3 \\ 9 &= 9 \end{align}\right. ✓ $$

The solution is (0, −3), and the system is consistent and independent because there is one solution.

Example 2: Solve a Linear System

Solve \(\left\{\begin{align} 2x + 3y &= -6 \\ 3x + 5y &= -11 \end{align}\right.\) by elimination.

Solution

The system is already written in columns. If the first equation were multiplied by 3 and the second equation were multiplied by −2, then the coefficients of x would be 6 and −6.

$$ \left\{\begin{align} 3\left(2x + 3y\right) &= 3\left(-6\right) \\ -2\left(3x + 5y\right) &= -2\left(-11\right) \end{align}\right. $$

$$ \left\{\begin{align} 6x + 9y &= -18 \\ -6x - 10y &= 22 \end{align}\right. $$

Add the equations

$$ \begin{align} 6x + 9y &= -18 \\ +\quad \underline{-6x - 10y} &= \underline{22} \\ -y &= 4 \end{align} $$

$$ y = -4 $$

Back-substitute this into one of the original equations.

$$ 2x + 3y = -6 $$

$$ 2x + 3\left(-4\right) = -6 $$

$$ 2x - 12 = -6 $$

$$ 2x = 6 $$

$$ x = 3 $$

The solution seems to be (3, −4). Check this by plugging it into both of the original equations.

$$ \left\{\begin{align} 2x + 3y &= -6 \\ 3x + 5y &= -11 \end{align}\right. $$

$$ \left\{\begin{align} 2(3) + 3(-4) &= -6 \\ 3(3) + 5(-4) &= -11 \end{align}\right. $$

$$ \left\{\begin{align} -6 &= -6 \\ -11 &= -11 \end{align}\right. ✓ $$

The solution is (3, −4), and the system is consistent and independent because there is one solution.

Try It 1

Solve \(\left\{\begin{align} -3x + 5y &= -21 \\ 2x - 7y &= 25 \end{align}\right.\) by elimination.

Answer

(2, −3); consistent and independent

Example 3: Solve a Linear System

Solve \(\left\{\begin{align} 2x - 6y = 4 \\ x = 3y + 2 \end{align}\right.\) by elimination.

Solution

The system is not written in columns. In the second equation, subtract 3y from both sides to make the columns line up.

$$ \left\{\begin{align} 2x - 6y &= 4 \\ x - 3y &= 2 \end{align}\right. $$

If the second equation were multiplied by -2, then the coefficients of x would be 2 and -2.

$$ \left\{\begin{align} 2x - 6y &= 4 \\ -2\left(x - 3y\right) &= -2\left(2\right) \end{align}\right. $$

$$ \left\{\begin{align} 2x - 6y &= 4 \\ -2x + 6y &= -4 \end{align}\right. $$

Add the equations

$$ \begin{align} 2x - 6y &= 4 \\ +\quad \underline{-2x + 6y} &= \underline{-4} \\ 0 &= 0 \end{align} $$

All the variables were eliminated and a true statement resulted. Thus there are infinitely many solutions described by the equation \(x - 3y = 2\), and the system is consistent and dependent.

Try It 2

Solve \(\left\{\begin{align} 14x + 21y &= 35 \\ -6x - 9y &= -12 \end{align}\right.\) by elimination.

Answer

No solution; inconsistent

Example 4: Problem Solving

An instrument company makes guitars and is starting production on a new model. It has a one-time cost of $10,050 to set up the factory production line and materials cost $150 per guitar. The company is going to sell the the guitars for $300 each. How many guitars do they need to sell to break even, where the costs equal the revenue?

Solution

Write two equations: one for cost on one for revenue. The costs are \(C = 10,050 + 150x\) where x is the number of guitars made. The revenue is the amount of money earned, \(R = 300x\). This question asks for the break even point where cost equals revenue. Replace the C and R with y since they are equal, and write the system of equations.

$$ \left\{\begin{align} y &= 10,050 + 150x \\ y &= 300x \end{align}\right. $$

Rewrite to make nice columns.

$$ \left\{\begin{align} 150x - y &= -10,050 \\ 300x - y &= 0 \end{align}\right. $$

Multiply the top equation by −1 and add.

$$ \begin{align} -150x + y &= 10,050 \\ +\quad \underline{300x - y} &= \underline{0} \\ 150x &= 10,050 \end{align} $$

Solve.

$$ x = 67 $$

The company would need to sell 67 guitars to break even.

Lesson Summary
Type of Solutions of Systems
1 solution
1 Solution: Consistent and Independent
many solutions
Infinitely Many Solutions: Consistent and Dependent
no solution
No Solution: Inconsistent
Solving Linear Equations by Elimination
  1. Write the equations in columns (ax + by = c).
  2. Obtain coefficients of one variable that differ only in sign by multiplying the equations by constants.
  3. Add the equations and solve the resulting equation. (A variable will be eliminated.)
  4. Back-substitute the answer into either original equation and solve.
  5. Check your solution.

If all the variables are eliminated and the result is

Helpful videos about this lesson.

Practice Exercises

    Check to see if the given point is a solution to the system.

  1. \(\left\{\begin{align} x + 6y &= -5 \\ 3x + 2y &= 1 \end{align}\right.\); (1, -1)
  2. \(\left\{\begin{align} 2x - y &= \frac{11}{2} \\ 3x + 2y &= 9 \end{align}\right.\); \(\left(3, \frac{1}{2}\right)\)

    3. Solve the system of equations and classify.

  3. \(\left\{\begin{align} x + 4y &= 0 \\ 3x - y &= 13 \end{align}\right.\)
  4. \(\left\{\begin{align} 7x - 5y &= -35 \\ 2x + 2y &= 14 \end{align}\right.\)
  5. \(\left\{\begin{align} 4x + 9y &= -4 \\ -2x - 6y &= 3 \end{align}\right.\)
  6. \(\left\{\begin{align} 5x - 3y &= 11 \\ -4x + \frac{12}{5}y &= \frac{41}{5} \end{align}\right.\)
  7. \(\left\{\begin{align} y &= -3x + 2 \\ x &= -\frac{1}{2}y - \frac{1}{2} \end{align}\right.\)
  8. \(\left\{\begin{align} 17x + 34y &= 2 \\ -51x - 68y &= -9 \end{align}\right.\)
  9. \(\left\{\begin{align} \frac{2}{3}x - \frac{5}{3}y &= \frac{7}{3} \\ y &= 0.4x - 1.4 \end{align}\right.\)
  10. \(\left\{\begin{align} -15x + 16y &= 29 \\ 5x - 12y &= -18 \end{align}\right.\)

    11. Problem Solving

  11. The Old Testament specified that people had to sacrifice a lamb at the temple for forgiveness of sins, but if they were poor, people could sacrifice a pair of doves. Two groups of travelers went to the temple and needed to purchase their sacrifices. The first group purchased 2 lambs and 3 pairs of doves for a total of $50.70 in today's dollars. The second group purchased 4 lambs and 1 pair of doves for $81.90. The groups tried to find out how much they were charged per lamb and pair of doves.
    1. How much were the people charged per lamb and pair of doves?
    2. Look up Mark 11:15-19. What did Jesus do when he saw this?
    3. What should the temple have been?
  12. Sally sells 10 shells at the seashore. A tourist paid her $2 for each perfect shell and $0.50 for each broken shell. If Sally received $11, how many of each type of shell did Sally collect and sell?
  13. Jill wants to make 10 L of 20% bleach solution by mixing some 10% solution and some 50% solution. How much of each type of solution should she use?
  14. The soccer club has two fee plans. Plan A is a $100 member fee and $5 per game you play. Plan B is no member fee but $15 per game you play. How many games will you have to play for both plans to cost the same and how much will that cost?
  15. Johnny invests $500 in two accounts that earn 1% and 0.5% interest. If he earns to $4.25 in interest, how much did he deposit in the accounts?

    16. Mixed Review

  16. (8-01) Solve by substitution: \(\left\{\begin{align} x + y = 7 \\ y = x^2 + 1 \end{align}\right.\)
  17. (8-01) Solve by graphing: \(\left\{\begin{align} x - y = 1 \\ y = \frac{1}{x - 1} \end{align}\right.\)
  18. (7-09) Write a polar equation of an ellipse with \(e = \frac{1}{3}\) and directrix \(x = -5\).
  19. (7-07) Graph the polar coordinates: \(A\left(4, \frac{2\pi}{3}\right)\) and \(B\left(-3, \frac{3\pi}{2}\right)\).
  20. (6-05) Evaluate \(\langle 1, -4 \rangle \cdot \langle 6, 3 \rangle\).

Answers

  1. Yes
  2. No
  3. (4, -1)
  4. (0, 7)
  5. \(\left(\frac{1}{2}, -\frac{2}{3}\right)\)
  6. No solution
  7. (3, -7)
  8. \(\left(\frac{5}{17}, -\frac{3}{34}\right)\)
  9. Many solutions
  10. \(\left(-\frac{3}{5}, \frac{5}{4}\right)\)
  11. Lamb: $19.50, Doves: $3.90; you look it up
  12. 4 perfect, 6 broken
  13. 2.5 L of 50%, 7.5 L of 10%
  14. 10 games, $150
  15. $350 at 1%, $150 at 0.5%
  16. (−3, 10), (2, 5)
  17. (0, −1), (2, 1)
  18. \(r = \frac{5}{3 - \cos \theta}\)
  19. answer
  20. −6