Solve the Systems of Equations by the Substitution Method

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Exercise 1

\left\{\begin{matrix} 3x - 4y = -6 \\ 2x + 4y = 16 \end{matrix}\right

Exercise 2

\left\{\begin{matrix} 2x + 3y = -1 \\ 3x + 4y = 0 \end{matrix}\right

Exercise 3

\left\{\begin{matrix} \frac { x + 3y }{ 2 } = 5 \\ 3x - y = 5y \end{matrix}\right

Exercise 4

\left\{\begin{matrix} \frac { x + 3y }{ 2 } = 5 \\ 4 - \frac { 2x - y }{ 2 } = 1 \end{matrix}\right

Exercise 5

\left\{\begin{matrix} x + y = 60 \\ 16x + 20y = 1100 \end{matrix}\right

Exercise 6

\left\{\begin{matrix} 3x + 2y = 7 \\ 4x - 3y = -2 \end{matrix}\right

 

Solution of exercise 1

\left\{\begin{matrix} 3x - 4y = -6 \\ 2x + 4y = 16 \end{matrix}\right

2x = 16 - 4y \qquad x = 8 - 2y

 

3x - 4y = -6

3(8 - 2y) - 4y = -6

24 - 6y - 4y = -6

24 - 10y = -6

- 10y = -30

y = 3

 

x = 8 - 2(3)

x = 8 - 6

x = 2

 

Solution of exercise 2

\left\{\begin{matrix} 2x + 3y = -1 \\ 3x + 4y = 0 \end{matrix}\right

3x = -4y \qquad x = \frac { -4y }{ 3 }

Plugging the value of x in the first equation:

2x + 3y = -1

2(\frac { -4y }{ 3 }) + 3y = -1

\frac { -8y }{ 3 } + 3y = -1

-8y + 9y = -3

y = -3

 

x = \frac { -4 (-3) }{ 3 }

x = 4

 

Solution of exercise 3

\left\{\begin{matrix} \frac { x + 3y }{ 2 } = 5 \\ 3x - y = 5y \end{matrix}\right

\left\{\begin{matrix} x + 3y = 10 \\ 3x = 6y \end{matrix}\right

3x = 6y \qquad x = 2y

Plugging the value fo x in the first equation:

x + 3y = 10

2y + 3y = 10

5y = 10

y = 2

 

Plugging the value of y in the second equation:

x = 2y

x = 2(2)

x = 4

 

Solution of exercise 4

\left\{\begin{matrix} \frac { x + 3y }{ 2 } = 5 \\ 4 - \frac { 2x - y }{ 2 } = 1 \end{matrix}\right

\left\{\begin{matrix} x + 3y = 10 \\ \frac { 8 - 2x - y }{ 2 } = 1 \end{matrix}\right

\left\{\begin{matrix} x = 10 - 3y \\   -2x + y = -6 \end{matrix}\right

Plugging the first equation in the second equation:

-2x + y = -6

-2(10 - 3y) + y = -6

-20 + 6y + y = -6

-20 + 6 = -7y

7y = 14

y = 2

 

x = 10 - 3y

x = 10 - 3(2)

x = 10 - 6

x = 4

 

Solution of exercise 5

\left\{\begin{matrix} x + y = 60 \\ 16x + 20y = 1100 \end{matrix}\right

x = 60 - y

Plugging the value of x in the second equation:

16x + 20y = 1100

16(60 - y) + 20y = 1100

960 - 16y +20y = 1100

4y = 1100 - 960

4y = 140

y = 35

 

x = 60 - y

x = 60 - 35

x = 25

 

Solution of exercise 6

\left\{\begin{matrix} 3x + 2y = 7 \\ 4x - 3y = -2 \end{matrix}\right

3x = 7 - 2y \qquad x = \frac { 7 - 2y }{ 3 }

Plugging the value of x in the second equation:

4x - 3y = -2

4(\frac { 7 - 2y }{ 3 }) - 3y = -2

\frac { 28 - 8y }{ 3 } - 3y = -2

\frac { 28 - 8y - 9y }{ 3 } = -2

28 - 17y = -6

-17y = -34

y = 2

 

x = \frac { 7 - 2(2) }{ 3 } = \frac { 7 - 4 }{ 3 }

x = 1

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.