MAT 101 GRADED EXAM 4 TOPIC 14 2026
PRACTICE QUESTIONS AND SOLUTIONS
GUARANTEED PASS
◉ We also categorize the equations in a system of equations by
calling the equations *independent* or *dependent*. What is the
difference between independent equations and dependent
equations. Answer: If two equations are independent, they each have
their own set of solutions. Intersecting lines and parallel lines are
independent.
If two equations are dependent, all the solutions of one equation are
also solutions of the other equation. When we graph two dependent
equations, we get coincident lines.
◉ EXAMPLE :
Without graphing, determine the number of solutions and then
classify the system of equations.
ⓐ y = 3x − 1 ⓑ 2x + y = − 3
6x − 2y = 12 x − 5y = 5. Answer: Solution:
ⓐ We will compare the slopes and intercepts of the two lines.
,(1) So you need to rewrite both of the equations in slope-intercept
form:
The first equation is already in slope-intercept form: y = 3x - 1
6x - 2y = 12 ==> y = 3x − 6
(2) Find the slope and intercept of each line.
y = 3x − 1 y = 3x − 6
m=3m=3
b = −1 b = −6
Since the slopes are the same and y-intercepts are
different, the lines are parallel.
*A system of equations whose graphs are parallel lines has no
solution and is inconsistent and independent*
ⓑ (Use the same steps above)
*A system of equations whose graphs are intersected, has 1 solution
and is consistent and independent*
◉ HOW TO SOLVE A SYSTEM OF EQUATIONS BY *SUBSTITUTION*:.
Answer: (1) Solve one of the equations for *either* variable.
(2) Substitute the expression from Step 1 into the other equation.
(3) Solve the resulting equation.
, (4) Substitute the solution in Step 3 into either of the original
equations to find the other variable.
(5) Write the solution as an *ordered pair*.
(6) Check that the ordered pair is a solution to *both* original
equations.
◉ HOW TO SOLVE A SYSTEM OF EQUATIONS BY *ELIMINATION*:.
Answer: (1) Write both equations in *standard form*. If any
coefficients are fractions, clear them.
(2) Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that
variable are opposites.
(3) Add the equations resulting from Step 2 to eliminate one
variable.
(4) Solve for the remaining variable.
(5) Substitute the solution from Step 4 into one of the original
equations. Then solve for the other variable.
(6) Write the solution as an ordered pair.
(7) Check that the ordered pair is a solution to *both* original
equations
◉ *standard form* of a linear equation. Answer: Ax + By = C, where
A,B, and C are real numbers, and A and B are not both zero.
PRACTICE QUESTIONS AND SOLUTIONS
GUARANTEED PASS
◉ We also categorize the equations in a system of equations by
calling the equations *independent* or *dependent*. What is the
difference between independent equations and dependent
equations. Answer: If two equations are independent, they each have
their own set of solutions. Intersecting lines and parallel lines are
independent.
If two equations are dependent, all the solutions of one equation are
also solutions of the other equation. When we graph two dependent
equations, we get coincident lines.
◉ EXAMPLE :
Without graphing, determine the number of solutions and then
classify the system of equations.
ⓐ y = 3x − 1 ⓑ 2x + y = − 3
6x − 2y = 12 x − 5y = 5. Answer: Solution:
ⓐ We will compare the slopes and intercepts of the two lines.
,(1) So you need to rewrite both of the equations in slope-intercept
form:
The first equation is already in slope-intercept form: y = 3x - 1
6x - 2y = 12 ==> y = 3x − 6
(2) Find the slope and intercept of each line.
y = 3x − 1 y = 3x − 6
m=3m=3
b = −1 b = −6
Since the slopes are the same and y-intercepts are
different, the lines are parallel.
*A system of equations whose graphs are parallel lines has no
solution and is inconsistent and independent*
ⓑ (Use the same steps above)
*A system of equations whose graphs are intersected, has 1 solution
and is consistent and independent*
◉ HOW TO SOLVE A SYSTEM OF EQUATIONS BY *SUBSTITUTION*:.
Answer: (1) Solve one of the equations for *either* variable.
(2) Substitute the expression from Step 1 into the other equation.
(3) Solve the resulting equation.
, (4) Substitute the solution in Step 3 into either of the original
equations to find the other variable.
(5) Write the solution as an *ordered pair*.
(6) Check that the ordered pair is a solution to *both* original
equations.
◉ HOW TO SOLVE A SYSTEM OF EQUATIONS BY *ELIMINATION*:.
Answer: (1) Write both equations in *standard form*. If any
coefficients are fractions, clear them.
(2) Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that
variable are opposites.
(3) Add the equations resulting from Step 2 to eliminate one
variable.
(4) Solve for the remaining variable.
(5) Substitute the solution from Step 4 into one of the original
equations. Then solve for the other variable.
(6) Write the solution as an ordered pair.
(7) Check that the ordered pair is a solution to *both* original
equations
◉ *standard form* of a linear equation. Answer: Ax + By = C, where
A,B, and C are real numbers, and A and B are not both zero.
