Problems Involving Linear Functions and Systems of Linear Equations

3.11 Problems Involving Linear Functions

Objective:

• Use linear functions to solve problems.

Steps in Problem Solving

1. Read and analyze the problem.
2. Make or devise a plan.
   • Determine what is asked in the problem.
   • Decide what to do.
3. Carry out the plan (Solve the problem).
4. Check your answer.

Number Problems

• Considered the easiest because the numbers are directly stated.

Example:

• A number is four more than twice another. Their sum is 13. What are the numbers?

Representation of the unknown:

• Let $x$ = the first number
• Let $y$ = the other number

Translation and derivation of the equation:

• $x + y = 13$ (their sum is 13)
• Solve for $x$ and $y$
• $x + y = 13$
• $2y + 4 + y = 13$ (since the number is four more than twice another)
• $3y + 4 = 13$ (combine similar terms)
• $3y = 13 – 4$ (transpose 4 to the right side)
• $3y = 9$ (simplify)
• $y = 3$
• Therefore, the other number is 3 and the first number is $2y + 4 = 2(3) + 4 = 10$.

Check the solution:

• $y + 2y + 4 = 13$
• $3 + 2(3) + 4 = 13$
• $13 = 13$

Age Problems

• Word problems about the current, past, or future ages of people.

Example:

• Odette’s age in October 2012 was one more than twelve times as old as her daughter Samara. In October 2017, her age will be two more than four times as old as her daughter. How old is her daughter?

Representation of the unknown:

• Let $x$ = the age of the daughter in 2012
• $12x + 1$ = the age of the mother in 2012

Translation and derivation of equations:

• Using the second relationship: In 2017, her age will be two more than four times as her daughter, we have;
• $12x + 1 + 5 = 4(x + 5) + 2$
• $12x + 6 = 4x + 22$
• $12x – 4x = 22 – 6$
• $8x = 16$
• $x = 2$ (age of daughter)
• $12(2) + 1 = 25$ (age of mother)

Check the solution:

• Odette’s age is one more than twelve times as old as her daughter
• $12x + 1 = 25$
• $12(2) + 1 = 25$
• $24 + 1 = 25$
• $25 = 25$
• Her age will be two more than four times as old as her daughter in 2017 (after 5 years)
• $12x + 1 + 5 = 4(x + 5) + 2$
• $12(2) + 6 = 4(2 + 5) + 2$
• $24 + 6 = 4(7) + 2$

Systems of Linear Equations

Objective:

1. Identify the kinds of systems of linear equations by comparing the ratios of coefficients of $x$, $y$, and constant, given a pair of linear equations in two variables.
2. Identify the kinds of systems of linear equations by comparing slope and y-intercepts, given a pair of linear equations in two variables.

Systems of Equations

• The process of solving two or more variables by using two or more equations.
• The solution set of a linear system of equations contains all ordered pairs that satisfy all the equations of a system at the same time.

Kinds of Systems

1. Independent or consistent

• Without graphing, if the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$ and also not equal to the ratio of constants, then the system is independent or consistent or has a single solution.
• In symbols; $\frac{a<em>1}{a</em>2} ≠ \frac{b<em>1}{b</em>2} ≠ \frac{c<em>1}{c</em>2}$
• If the slopes are different, then the system is consistent with independent equations and has a single solution.

Example:

• Eq1. = $x + 2y = 4$
• Eq2 = $6x – 2y = 3$

Solution:

• Eq1 = $x + 2y = 4$
• Eq2= $6x – 2y = 3$
• $x + 2y – 4 = 0$ $a=1, b=2, c=-4$
• $6x – 2y – 3 = 0$ $a=6,b=-2, c=-3$
• $\frac{1}{6} ≠ \frac{2}{-2} ≠ \frac{-4}{-3}$
• $\frac{a<em>1}{a</em>2} ≠ \frac{b<em>1}{b</em>2} ≠ \frac{c<em>1}{c</em>2}$
• Independent or consistent, has one solution.

For the Slope intercept Form

• $x + 2y = 4$
• $6x – 2y = 3$
• $2y = - x + 4$
• $2y = 6x – 3$
• $\frac{2y}{2} = - \frac{x}{2} + \frac{4}{2}$
• $\frac{2y}{2} = - \frac{6x}{2} + \frac{3}{2}$
• $y = - \frac{1}{2}x + 2$ $m = - \frac{1}{2}, y = 2$
• $y = - 3x + \frac{3}{2}$ $m = - 3, y = \frac{3}{2}$
• The slopes ($m$) and y-intercepts ($b$) are different.
• Independent or consistent, has one solution.

2. Inconsistent

• Without graphing, if the ratio of the coefficients of $x$ is equal to the ratio of the coefficients of $y$ but not equal to the ratio of constants, then the system is inconsistent.
• In symbols: $\frac{a<em>1}{a</em>2} = \frac{b<em>1}{b</em>2} ≠ \frac{c<em>1}{c</em>2}$
• If the slopes are equal with different y-intercepts, then the system is inconsistent and has no solution.

Example:

• $7x – y = 1$
• $7x – y = 47$

Solution:

• Eq1 = $7x – y = 1$
• Eq2= $7x – y = 47$
• $7x – y -1 = 0$ $a=7, b= - 1, c= - 1$
• $7x – y - 47= 0$ $a= 7,b=- 1, c=- 47$
• $\frac{7}{7} = \frac{-1}{-1} ≠ \frac{-1}{-47}$
• $\frac{a<em>1}{a</em>2} = \frac{b<em>1}{b</em>2} ≠ \frac{c<em>1}{c</em>2}$
• Independent or Inconsistent, has no solution.

For the Slope intercept Form

• $7x – y = 1$
• $7x – y = 47$
• $y = 7x - 1$ $m = 7 , y = -1$
• $y = 7x - 47$ $m = 7 , y = -47$
• The slopes ($m$) are the same and y-intercepts ($b$) are different.
• Independent or consistent, has no solution.

3. Dependent

• Without graphing, if the ratio of the coefficient x is equal to the ratio of the coefficients of y and equal to the ratio of constants, then the system is dependent.
• In symbols; $\frac{a<em>1}{a</em>2} = \frac{b<em>1}{b</em>2} = \frac{c<em>1}{c</em>2}$
• If the slopes and y-intercepts are equal, then the system is consistent with dependent equations and has an infinite number of solutions.

Example:

• Eq1. $x + 2y = 4$
• Eq2. $2x + 4y = 8$

Solution:

• Eq1 = $x + 2y = 4$
• Eq2= $2x + 4y = 8$
• $x + 2y – 4 = 0$ $a=1, b=2, c=-4$
• $2x + 4y -8 = 0$ $a=1,b=2, c=-4$
• $\frac{1}{1} = \frac{2}{2} = \frac{4}{4}$
• $\frac{a<em>1}{a</em>2} = \frac{b<em>1}{b</em>2} = \frac{c<em>1}{c</em>2}$
• Dependent, has many solutions.

For the Slope intercept Form

• $x + 2y = 4$
• $2x + 4y = 8$
• $2y = - x + 4$
• $4y = -2x + 8$
• $\frac{2y}{2} = - \frac{x}{2} + \frac{4}{2}$
• $\frac{4y}{4} = - \frac{-2x}{4} + \frac{8}{4}$
• $y = - \frac{1}{2}x + 2$ $m = - \frac{1}{2} , y = 2$
• $y = - \frac{-1x}{2} + 2$ $m = - \frac{1}{2} , y =2$
• The slopes ($m$) and y-intercepts ($b$) are different.
• Independent or consistent, has one solution.

Solving Systems of Linear Equations by Substitution

Objective:

1. Find the solution of systems of linear equations in two variables by substitution.

Lesson Proper

• To find the solution of a system of linear equations in two variables is to look for ($x$, $y$) that will satisfy both equations in the system.
• One method in finding the solution of a system of linear equations algebraically is the method of substitution.

Steps in solving system of linear equations in two variables by substitution method.

1. Find the value of either unknown in terms of the other.
2. Substitute the expression obtained in the other equation.
3. Solve that equation,
4. Find the corresponding value of the other variable.
5. Check the solution in both of the original equation.

Example 1

• Solve the system:
• $x + 3y = 12$
• $y = x – 4$

Solution:

1. Substitute $x – 4$ for $y$ in the first equation, and solve for $x$.

   • $x + 3y = 12$
   • $x + 3 (x – 4) = 12$
   • $x + 3x – 12 = 12$
   • $4x – 12 = 12$
   • $4x = 12 + 12$
   • $4x = 24$
   • $x = 6$

2. Substitute this value of $x$ in any of the equation.

3. Check if $(6, 2)$ satisfies both of the original equations.

Example 2

• Solve the system:
• $x – 5y = 9$
• $x – 2y = 6$

Solution:

1. Solve the second equation for $x$ in terms of $y$.

   • $x – 2y = 6$
   • $x = 2y + 6$

2. Substitute this expression for $x$ in the first equation, and solve for $y$.

   • $x – 5y = 9$
   • $2y + 6 – 5y = 9$
   • $-3y + 6 = 9$
   • $-3y = 9 – 6$
   • $-3y = 3$
   • $y = -1$

3. Substitute this value of $y$ in any of the equation.