Precalculus by Richard Wright

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

Summary: In this section, you will:

- Solve two-variable linear systems using elimination.
- Classify types of solutions.

SDA NAD Content Standards (2018): PC.6.1

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 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.

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

If all the variables are eliminated and the result is

- True (like 0 = 0), there are infinitely many solutions.
- False (like 0 = 9), there are no solutions.

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

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.

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

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.

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

(2, −3); consistent and independent

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

The system is not written in columns. In the second equation, subtract 3*y* 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.

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

No solution; inconsistent

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?

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.

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

If all the variables are eliminated and the result is

- True (like 0 = 0), there are infinitely many solutions.
- False (like 0 = 9), there are no solutions.

Helpful videos about this lesson.

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

- How much were the people charged per lamb and pair of doves?
- Look up Mark 11:15-19. What did Jesus do when he saw this?
- What should the temple have been?

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

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

Solve the system of equations and classify.

Problem Solving

Mixed Review

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