13 system of equations

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SYSTEMS OF
EQUATIONS
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Janice Levasseur
System of Equations
• A system of equations is two or more
equations to be solved simultaneously
• A solution of a system of equations is a
solution that makes each equation true
• We will work with a system of two
equations with two unknowns (variables)
Ex: Determine if the point (1,5) is a
solution to the system
5x – 2y = -5
3x – 7y = - 32
• Is the point a solution to the first equation?
• Check by plugging in x = 1 and y = 5
 5(1) – 2(5) = 5 – 10 = - 5 YES
• Is the point a solution to the second equation?
• Check by plugging in x = 1 and y = 5
 3(1) – 7(5) = 3 – 35 = - 32 YES
• Therefore, (1,5) is a solution to the system.
Ex: Determine if the point (4,2) is a
solution to the system
3y – 2x = - 2
y + 2x = 8
• Is the point a solution to the first equation?
• Check by plugging in x = 4 and y = 2 (careful!)
 3(2) – 2(4) = 6 – 8 = - 2 YES
• Is the point a solution to the second equation?
• Check by plugging in x = 4 and y = 2 (careful!)
 2 + 2(4) = 2 + 8 = 10 NO
• Therefore, (4,2) is not a solution to the system.
Recall linear equations y = mx + b
• The graph of the line is the solution set.
(i.e. every point on the line is a solution to the
equation y = mx + b)
 Ex: Graph y = 2x – 1
Plugging in values for x we find three solutions
• (0, -1)
• (-1, -3)
• (1, 1)
(Remember, there are an infinite number of
possible solutions)
Plotting the points we get
Graph y = 2x – 1
(0, -1)
(-1, -3)
(1, 1)
Recall: every point on the line is a solution to
the equation y = 2x - 1
Plotting two lines
on the same set of axes
• If we were to plot two lines on the same
set of axes, the lines will intersect either
one time
no times (i.e. never intersect)
an infinite number of times
• The point(s) of intersection is/are the
solution(s) to the system of equations
Ex: Plot on the same set of axes
y = (3/2)x + 1 & y = (-1/4)x – (5/2)
• Picking “nice” values of x we find the
following three solutions for the 1st and 2nd
equations:
y = (3/2)x + 1
• (-2, -2), (0, 1), and (2, 4)
y = (-1/4)x – (5/2)
• (-4, -3/2), (0, -5/2), and (4, -7/2)
• Plot each line on the same set of axes
Graph y = (3/2)x + 1
Graph y = -(1/4)x – (5/2)
(-2, -2)
(-4, -3/2)
(0, 1)
(0, -5/2)
(2, 4)
(4, -7/2)
the lines intersect one time at (-2, -2)
 the solution to the system of
equations is (-2, -2)
The system of equations is Independent
Ex: Plot on the same set of axes
y = x + 2 and y = x – 1
• Picking values of x we find the following
three solutions for the 1st and 2nd equations:
y=x+2
• (-1, 1), (0, 2), and (1, 3)
y=x–1
• (-1, -2), (0, -1), and (1, 0)
• Plot each line on the same set of axes
Graph y = x + 2
Graph y = x – 1
(-1, 1)
(-1, -2)
(0, 2)
(0, -1)
(1, 3)
(1, 0)
the lines are parallel,
therefore they never intersect
there is no solution to the system of equations
The system of equations is Inconsistant
Ex: Plot on the same set of axes
y = (1/3)x + 2 and –x + 3y = 6
• Picking “nice” values of x we find the following
three solutions for the 1st and 2nd equations:
y = (1/3)x + 2
• (-3, 1), (0, 2), and (3, 3)
–x + 3y = 6
• (-3, 1), (0, 2), and (3, 3)
Graph y = (1/3)x + 2
Graph –x + 3y = 6
(-3, 1)
(-3, 1)
(0, 2)
(0, 2)
(3, 3)
(3, 3)
the lines are identical
therefore intersect at every point on the lines
there is an infinite number of solution
to the system of equations
The system of equations is Dependent
Types of Systems
• A system of equations with one solution
(the lines intersect one time) is said to be
independent
• A system of equations with no solutions
(the lines are parallel) is said to be
inconsistent
• A system of equations with an infinite
number of solutions (the lines are the
same) is said to be dependent
Ex: Graph the system to solve:
x–y=2
x+y=6
• Using the slope-intercept form of the
equations we find
• Equation 1: x – y = 2  y = x – 2
 m = 1 and b = - 2
• Equation 2: x + y = 6  y = - x + 6
 m = - 1 and b = 6
• Graph Eq 1: y = x – 2 m = 1 and b = - 2
• Graph Eq 2: y = - x + 6 m = - 1 and b = 6
Start on the y-axis at 6
Using m = -1  down 1, right 1
Start on the y-axis at -2
Using m = 1  up 1, right 1
The lines intersect one time at the point (4, 2)
 (4, 2) is the solution to the system
Your turn to try a few
An Algebraic Solution to a
System of Equations
• The graphical method to solve a system of
equations helps us understand what the
solution of a system is.
• However, unless that intersection occurs
at a “nice point” it will be difficult to
pinpoint the solution.
• We need an algebraic approach to solving
a system of equations.
The Substitution Method
The Substitution Method is an algebraic approach
to solving a system of equations.
1. In one of the equations, solve for one of the
variables.
2. Plug the expression for that variable into the
other equation and simplify. You now have an
equation in a single variable.
3. Solve the single-variable equation for the
remaining variable.
4. Evaluate the expression found in Step 1 using
the value of the variable found in Step 3 to
solve for the other variable.
5. Write the solution as an ordered pair,
alphabetically, and check
Ex: Solve the system
y = x – 6  Equation 1
x + y = - 2  Equation 2
Step 1: Solve one equation for one variable.
DONE, Equation 1:
y=x–6
Step 2: Plug the expression for that variable into
the other equation and simplify.
 x + (x – 6) = - 2
 2x – 6 = - 2
(adding 6 to both sides)  2x = 4
(dividing both sides by 2)
x =2
Equation 2: x + y = - 2
Ex: Solve the system
y=x–6
x+y=-2
x =2
Step 4: Evaluate the expression found in Step 1
using the value of the variable found in Step 3 to
solve for the other variable.
Equation 1: y = x - 6  y = 2 - 6
 y=–4
Step 5: Write the solution as an ordered pair & check.
Solution: (2, - 4)
Check: Equation 1: y = x – 6  - 4 = 2 – 6? YES!
Equation 2: x + y = - 2  2 + (-4) = - 2? YES!
Ex: Solve the system
x – y = - 3  Equation 1
2x + 3y = - 6  Equation 2
Step 1: Solve one equation for one variable.
Equation 1 solve for x: x = y – 3
Step 2: Plug the expression for that variable into
the other equation and simplify.
Equation 2: 2x + 3y= - 6
 2(y – 3) + 3y = - 6
 2y – 6 + 3y = - 6
 5y – 6 = - 6
Step 3: Solve the equation for the remaining variable.
5y – 6 = - 6
 5y = 0 (adding 6 to both sides)
 y = 0 (dividing both sides by 5)
Step 4: Evaluate the expression found in Step 1
using the value of the variable found in Step 3 to
solve for the other variable.
Step 1: x = y – 3
 x=0-3
 x=-3
Step 5: Write the solution as an ordered pair & check.
Solution: (- 3, 0)
Check: Equation 1: x – y = - 3  - 3 – 0 = - 3?
YES!
Equation 2: 2x + 3y = - 6  2(-3)+3(0) = - 6? YES!
Ex: Solve the system
y – 2x = – 6  Equation 1
2y - x = 5  Equation 2
Step 1: Solve one equation for one variable.
Equation 1 solve for y by adding 2x to both sides:
y = 2x – 6
Step 2: Plug the expression for that variable into
the other equation and simplify.
Equation 2: 2y – x = 5
 2(2x – 6) – x = 5
 4x – 12 – x = 5
 3x – 12 = 5
Step 3: Solve the equation for the remaining variable.
3x – 12 = 5
 3x = 17 (adding 12 to both sides)
 x = 17/3 (dividing both sides by 3)
Step 4: Evaluate the expression found in Step 1
using the value of the variable found in Step 3 to
solve for the other variable.
 y = 2(17/3) - 6  y = 16/3
Step 1: y = 2x - 6
Step 5: Write the solution as an ordered pair & check.
Solution: (17/3, 16/3)
Check: Equation 1:
y – 2x = – 6  16/3 – 2(17/3) = – 6? YES!
Equation 2: 2y – x = 5  2(16/3) – (17/3) = 5?
YES!
Your turn to try a few
The Elimination Method
The Elimination Method is another algebraic
approach to solving a system of equations.
1. Use the Multiplication Principle for Equations
to set up the system so that the coefficients of
one of the variables are opposites.
2. Add the equations to “eliminate” one of the
variables.
3. Solve the single-variable equation for the
remaining variable.
4. “Backsolve” – solve one of the original
equations for the other variable, using the
value of the first variable found in Step 3.
5. Write the solution as an ordered pair,
alphabetically, and check
Ex: Solve the system
x + y = 8  Equation 1
- x + 2y = 7  Equation 2
Step 1: set up the system so that the coefficients of
one of the variables are opposites.
Done. The coefficients of x are 1 and - 1
Step 2: Add the equations to “eliminate” one of the
variables.
Step 3: Solve the equation
 Equation 1
x+y =8
for the remaining variable.
-x + 2y = 7  Equation 2
(dividing both sides by 3)
y =5
3y = 15
Step 4: “Backsolve” – solve one of the original
equations for the other variable, using the value of
the first variable found in Step 3.
Equation 1: x + y = 8  x + 5 = 8
x=3
Step 5: Write the solution as an ordered pair & check.
Solution: (3, 5)
Check: Equation 1: x + y = 8  3 + 5 = 8?
YES!
Equation 2: - x + 2y = 7  - 3 + 2(5) = 7? YES!
Ex: Solve the system
3x – y = 8  Equation 1
x + 2y = 5  Equation 2
Step 1: set up the system so that the coefficients of one of the
variables are opposites.
The coefficients of y are opposite in sign: -1 and 2. If the -1
was a -2, then the coefficients of y would be opposites.
Multiply Equation 1 through by 2 to obtain an equivalent
equation: 2[3x – y = 8]  6x – 2y = 16.
Step 2: Add the equations to “eliminate” one of the variables.
6x – 2y = 16
x + 2y = 5
7x
= 21
 Equation 1 multiplied by 2
 Equation 2
x =3
Step 3: Solve the equation
for the remaining variable.
Step 4: “Backsolve”
Equation 1: 3x – y = 8
 3(3) – y = 8
9–y=8
y=1
Step 5: Write the solution as an ordered pair & check.
Solution: (3, 1)
Check: Equation 1: 3x – y = 8  3(3) - 1 = 8?
YES!
Equation 2: x + 2y = 5  3 + 2(1) = 5?
YES!
Ex: Solve the system
2w – 3z = - 1  Equation 1
3w + 4z = 24  Equation 2
Step 1: set up the system so that
the coefficients of one of the variables are opposites.
The coefficients of z are opposite in sign: -3 and 4.
Question: What is the LCM(-3, 4)? 12
Multiply Equation 1 through by 4 to obtain an equivalent
equation: 4[2w – 3z = -1]  8w – 12z = - 4.
Multiply Equation 2 through by 3 to obtain an equivalent
equation: 3[3w + 4z = 24]  9w + 12z = 72.
The coefficients of z are now opposites: - 12 and 12.
Step 2: Add the equations to “eliminate” one of the
variables.
8w – 12z = - 4
9w + 12z = 72
17w
= 68
 Equation 1 equivalent
 Equation 2 equivalent
Step 3: Solve the equation for the remaining variable.
17w = 68
 w = 4 (dividing both sides by 17)
Step 4: “Backsolve” Equation 1: 2w – 3z = -1
 2(4) – 3z = -1
 8 – 3z = - 1
 - 3z = - 9
z=3
Step 5: Write the solution as an ordered pair (in
alphabetical order) & check.
Solution: (w, z)  (4, 3)
Check: Equation 1: 2w – 3z = - 1  2(4) – 3(3)= -1? YES!
Equation 2: 3w + 4z = 24  3(4) + 4(3)= 24? YES!
Your turn to try a few
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