Lesson 4.6: Chapter Summary and Review

New Concept

This chapter reviews key concepts for analyzing linear relationships. You'll connect solving single equations to solving systems of equations, using algebraic methods and graphs to find solutions and model real-world scenarios.

What’s next

Let's consolidate your knowledge. You will now tackle a series of review problems, practice quizzes, and challenge questions covering all of Chapter 4.

Property

If $a$, $b$, and $c$ are any numbers, then

Steps for Solving Linear Equations.

1. Use the distributive law to remove any parentheses.
2. Combine like terms on each side of the equation.
3. By adding or subtracting the same quantity on both sides of the equation, get all the variable terms on one side and all the constant terms on the other.
4. Divide both sides by the coefficient of the variable to obtain an equation of the form $x = a$.

Examples

• To simplify $5(x+3)$, you distribute the 5 to both $x$ and 3. This gives $5 \cdot x + 5 \cdot 3$, which simplifies to $5x+15$.
• To simplify $6(2a-4) + 7$, first distribute the 6 to get $12a - 24 + 7$. Then, combine the constant terms to get the final expression $12a - 17$.
• To solve the equation $3(y-2) = 9$, first distribute the 3 to get $3y - 6 = 9$. Add 6 to both sides to get $3y = 15$, then divide by 3 to find $y=5$.

Explanation

Think of this law as 'distributing' the term outside the parentheses to every term inside. It helps break down multiplication with a group of terms, turning one complex multiplication into several simpler ones.

Property

A pair of linear equations in two variables

considered together is called a system of linear equations. A solution is an ordered pair $(x, y)$ that satisfies each equation in the system. There are three types of linear systems:

1. Consistent and independent system. The graphs of the two lines intersect in exactly one point. The system has exactly one solution.
2. Inconsistent system. The graphs of the equations are parallel lines and hence do not intersect. An inconsistent system has no solutions.
3. Dependent system. The graphs of the two equations are the same line. A dependent system has infinitely many solutions.

Examples

• A consistent system like $y=x+2$ and $y=-x+4$ has one solution, $(1,3)$, where the two lines intersect.
• An inconsistent system like $y=3x+1$ and $y=3x+5$ has no solution. The lines have the same slope, so they are parallel and never cross.
• A dependent system like $x+y=5$ and $2x+2y=10$ has infinite solutions. The second equation is just the first one multiplied by 2, so they represent the same line.

Explanation

Think of each equation as a line on a graph. The system's solution is where these lines meet. They can cross at one point, run parallel and never touch, or be the exact same line, lying on top of each other.

Property

To Solve a System by Substitution.

1. Choose one of the variables in one of the equations. (It is best to choose a variable whose coefficient is 1 or $-1$.) Solve the equation for that variable.
2. Substitute the result of Step 1 into the other equation. This gives an equation in one variable.
3. Solve the equation obtained in Step 2. This gives the solution value for one of the variables. Substitute this value into the result of Step 1 to find the solution value of the other variable.

Examples

• For the system $y = 2x$ and $x+y=9$, substitute $2x$ for $y$ in the second equation to get $x+2x=9$. This simplifies to $3x=9$, so $x=3$. Then find $y$ by using $y=2(3)=6$. The solution is $(3,6)$.
• For $x-3y=4$ and $2x+y=13$, first solve the first equation for $x$ to get $x=3y+4$. Substitute this into the second equation: $2(3y+4)+y=13$. This becomes $6y+8+y=13$, so $7y=5$, and $y=\frac{5}{7}$. Finally, $x=3(\frac{5}{7})+4=\frac{43}{7}$.
• For $y = 2x-1$ and $4x-2y=2$, substitute $y$ into the second equation: $4x-2(2x-1)=2$. This simplifies to $4x-4x+2=2$, or $2=2$. Since this is always true, the system is dependent and has infinite solutions.

Explanation

This method involves solving one equation for a variable, then substituting that expression into the other equation. This 'replaces' one variable, leaving you with a single-variable equation that is much easier to solve.

Property

To Solve a System by Elimination.

1. Write each equation in the form $Ax + By = C$.
2. Decide which variable to eliminate. Multiply each equation by an appropriate constant so that the coefficients of that variable are opposites.
3. Add the equations from Step 2 and solve for the remaining variable.
4. Substitute the value found in Step 3 into one of the original equations and solve for the other variable.

When Using Elimination to Solve a System.

1. If combining the two equations results in an equation of the form $0x + 0y = k$ ($k \neq 0$), then the system is inconsistent.
2. If combining the two equations results in an equation of the form $0x + 0y = 0$, then the system is dependent.

Examples

• To solve $2x+y=5$ and $3x-y=10$, add the equations together. The $y$ terms cancel, leaving $5x=15$, so $x=3$. Substitute $x=3$ into the first equation: $2(3)+y=5$, which gives $6+y=5$, so $y=-1$. The solution is $(3,-1)$.
• To solve $x+2y=3$ and $2x+4y=7$, multiply the first equation by $-2$ to get $-2x-4y=-6$. Adding this to the second equation gives $0=1$. This is false, so the system is inconsistent and has no solution.
• To solve $3x-y=4$ and $-6x+2y=-8$, multiply the first equation by 2 to get $6x-2y=8$. Adding this to the second equation gives $0=0$. This is true, so the system is dependent and has infinite solutions.

Explanation

This method's goal is to 'eliminate' a variable by adding the two equations together. You may need to multiply one or both equations first, so the coefficients of one variable become opposites, like $3x$ and $-3x$.

Property

To find an equation for the line of slope $m$ passing through the point $(x_1, y_1)$, we use the point-slope formula

or

To Fit a Line through Two Points.

1. Compute the slope between the two points.
2. Substitute the slope and either point into the point-slope formula.

Examples

• To find the equation for a line with slope $m=4$ passing through $(1,3)$, use the formula $y-y_1 = m(x-x_1)$ to get $y-3 = 4(x-1)$.
• To find the equation for the line passing through $(2,5)$ and $(4,11)$, first calculate the slope: $m = \frac{11-5}{4-2} = \frac{6}{2} = 3$. Then use the point-slope formula with the point $(2,5)$: $y-5 = 3(x-2)$.
• A line passes through $(-2, 8)$ with a slope of $-3$. To write its equation in slope-intercept form, start with point-slope: $y-8 = -3(x - (-2))$. This becomes $y-8 = -3(x+2)$. Distribute to get $y-8 = -3x-6$, then add 8 to both sides: $y = -3x+2$.

Explanation

This formula is the most direct way to write a line's equation when you have its slope and any point it passes through. It's built directly from the definition of slope, $m = \frac{\text{rise}}{\text{run}}$.