Linear Equtions

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  • Assignment 10.1: Solving Systems of Linear Equations Complete the following problems. Be sure to show all work. 1) Determine whether the given ordered pair is a solution to the system. a. (1, 5), b. (2, –6),

    2) Solve each system by either substitution or the addition method. Identify which method you are using. a. b. c. d. e.

    3) Solve each system. a. b.

    4) Graph the solution set for each system of linear inequalities. a. b. c.

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College Algebra
Assignment Handout
Assignment 10.1: Solving Systems of Linear Equations
Complete the following problems. Be sure to show all work.
1) Determine whether the given ordered pair is a solution to the system.

2 x  3 y  13
 x  7 y  36

a. (1, 5), 

 x  4 y  26
 x y 8

b. (2, –6), 

2) Solve each system by either substitution or the addition method. Identify
which method you are using.

 2 x  y  1
 x  5 y  14

a. 

3x  2 y  2
 x  3 y  14

b. 

 5x  y  7
15 x  3 y  21

c. 

0.2 x  0.3 y  10
 20 x  30 y  25

d. 

 4 x  y  11
6 x  3 y  19

e. 

3) Solve each system.

 x yz 0

a.  x  y  z  6
 x  y  2 z  7

 2 x  4 y  z  7

b.  x  2 y  3z  2
 4 x  2 y  3z  14

4) Graph the solution set for each system of linear inequalities.

 y5
 x  2

a. 

 x y 6
 y  2x  8

b. 

College Algebra Handout Assignment 10.1, V2.4
© 2012 Pearson Education

Page 1 of 11

2/15/2018

College Algebra
Assignment Handout

 x0

y0
c. 
3x  4 y  12


College Algebra Handout Assignment 10.1, V2.4
© 2012 Pearson Education

Page 2 of 11

2/15/2018

College Algebra
Assignment Handout
Solution
Question 1

2 x  3 y  13
 x  7 y  36

a. (1, 5), 

Substituting in the first equation, we have:
LHS: 2(1) − 3(5) = 2 − 15 = −13
RHS: -13
The LHS = RHS, therefore the ordered pair is a solution to the equation.
Substituting in the second equation, we have:
LHS: 1 + 7(5) = 1 + 35 = 36
RHS: 36
The LHS = RHS, therefore the ordered pair is a solution to the equation.
Since the ordered pair (1, 5) satisfies both equations, it is a solution to
the system of equations.

 x  4 y  26
 x y 8

b. (2, –6), 

Substituting in the first equation, we have:
LHS: 2 − 4(−6) = 2 + 24 = 26
RHS: 26
The LHS = RHS, therefore the ordered pair is a solution to the equation.
Substituting in the second equation, we have:
LHS: 2 − 6 = −4
RHS: 8
The LHS = RHS, therefore the ordered pair is NOT a solution to the
equation.
Since the ordered pair (2, −6) does not satisfy both equations, it is NOT
solution to the system of equations.

College Algebra Handout Assignment 10.1, V2.4
© 2012 Pearson Education

Page 3 of 11

2/15/2018

College Algebra
Assignment Handout
Solution
Question 2

 2 x  y  1
 x  5 y  14

a. 

Using substitution method, we have:
Making y subject of the formula in the first equation we have:
𝑦 = 1 + 2𝑥

equation (3)

substituting equation (3) into equation (2), we have:
𝑥 = 5(1 + 2𝑥) − 14
𝑥 = 5 + 10𝑥 − 14
𝑥 − 10𝑥 = 5 − 14
−9𝑥 = −9
∴𝑥=1
substituting 𝑥 = 1 in equation (3), we have:
𝑦 = 1 + 2𝑥 = 1 + 2(1)
𝑦=3
Therefore, the solution to the system of equation is the ordered pair (𝟏, 𝟑)

3x  2 y  2
 x  3 y  14

b. 

Using substitution method, we have:
Making x subject of the formula in the second equation we have:
𝑥 = 14 + 3𝑦

equation (3)

substituting equation (3) into equation (1), we have:
3(14 + 3𝑦) + 2𝑦 = −2
42 + 9𝑦 + 2𝑦 = −2
9𝑦 + 2𝑦 = −2 − 42
11𝑦 = −44
∴𝑦=−

44
= −4
11

College Algebra Handout Assignment 10.1, V2.4
© 2012 Pearson Education...


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