Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations Exercise 5.4

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations

Page 147  Essential Question  Answer

Let’s consider a system of equations.
It is required to describe whether a system can have no solutions or infinitely many solutions.

Yes, a system of linear equations has no solution if the equations are inconsistent or the graph for the system has parallel lines.
A system has infinitely many solutions when it is consistent, and the graph for the system of equations coincides.

Yes, a system of linear equations can have no solution or infinitely many solutions.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations Exercise 5.4

Page 147  Exercise 2  Answer

Given: We are given the number of small and large beads in a necklace and a bracelet. We know the weight of the necklace and the bracelet.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 1

 

To find We have to write a system of linear equations that represents the situation.
We will assume “x” is the weight (in grams) of a small bead and “y” is the weight (in grams) of a large bead.
Then we will multiply the number with their corresponding weight and get the equation.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

“x” is the weight (in grams) of a small bead and “y” is the weight (in grams) of a large bead
The equation that represents the weight of the necklace is:   40x + 6y = 10
The equation that represents the weight of the bracelet is:  20x + 3y = 5

 

The system of equations that represent the given scenario is

​⇒ 40x + 6y = 10

​⇒ 20x + 3y = 5

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 2

 

Given: We are given the number of small and large beads in a necklace and a bracelet. We know the weight of the necklace and the bracelet.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 3

To find We have to graph the system in the coordinate plane and describe the two lines.
We will plot these graphs and check whether these lines are parallel or perpendicular or neither.

 

The system of equations is :

​⇒   40x + 6y = 10

​⇒   20x + 3y = 5

 

The plot of the system of equations is given below:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 4

The plot of these two equations is coincident.
They can be represented by the same line as shown above in the graph.

 

The plot of the system of equations is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 5

The two lines are coincidental.

 

Given: We are given the number of small and large beads in a necklace and a bracelet. We know the weight of the necklace and the bracelet.

To find We have to tell whether we can find the weight of each type of bead.
Now We calculate the weight of each bead by solving the system of equations.

We have made the system of equations in terms of the weight of each bead.
So when we solve them we can easily get to know the weight of each bead.

We can find the weight of each bead by solving the system of equations.
There are two equations and there are two variables in the equations. So, we can solve these equations for the values of “x” and “y”.

 

Page 148  Exercise 3  Answer

Given: A system of linear equations.

To find We have to tell whether a system of linear equations can have no solution and infinitely many solutions.
A system of equations can have a unique solution, no solution, or infinitely many solutions.
We get to know about the solution of a system of equations by looking at the coefficient of variables and the constant part.

 

An example of a system of equations with no solution:

​⇒  ​ 2x + 3y = 10

​⇒   2x + 3y = 12

The coefficients of “x” and “y” are the same but the constants in both the equations are different. So, this system will have no solution.

 

An example of a system of equations with infinitely many solutions:

​​⇒   2x + 3y=6

​⇒   4x + 6y=12

The coefficients of “x” and “y” and the constant are proportional. So, they basically represent the same line on the graph. They have infinitely many solutions.


Yes, a linear system of equations has no solution or infinitely many solutions. The example of no solution is ​ 2x + 3y = 10, 2x + 3y = 12. The example of infinitely many solutions is​  2x + 3y = 6, 4x + 6y = 12.

 

Page 150  Exercise 2  Answer

Given: A system of equations is given to us. It is, y = 5x−1 y = −5x + 5

To findWe have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

​y = 5x−1——-(1)

y =−5x + 5——(2)

 

After adding them, we get

​⇒   ​2y = 5x−5x−1 + 5

​⇒   2y = 4

​⇒   y = 2

Putting y=2 in equation (1) we get

​​⇒   2 = 5x−1

​⇒   3 = 5x

​⇒   x\(=\frac{3}{5}\)

The solution of the given system of equations y=5x−1,y=−5x+5 is x\(=\frac{3}{5}\) y=2.

 

Page 150  Exercise 3  Answer

Given: A system of equations is given to us. It is  2x−3y = 10, −2x + 3y = −10

​To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

​2x−3y = 10——-(1)

​−2x+3y = −10——(2)

After adding, Equation  (1) and  Equation (2), we get

​⇒   2x−2x−3y + 3y = 10−10

​⇒   0 = 0

This is true, so the given system of equations has infinitely many solutions.

The given system of equations 2x−3y = 10,−2x+3y = −10, has infinitely many solutions.

 

Page 150  Exercise 4  Answer

Given: A system of equations is given to us. It is   x + 3y = 6, −x−3y = 3

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

x+3y = 6——(1)

−x−3y = 3—–(2)

After adding  Equation (1) and  Equation  (2), we get

​⇒  x−x + 3y−3y = 6 + 3

​⇒  0 = 9

This is not true, so the solution of the given system of equations does not exist.

The solution of the given system of equations x + 3y = 6,−x−3y = 3, does not exist.

 

Page 150  Exercise 5  Answer

Given: A system of equations is given to us. It is  6x + 6y = −3, −6x−6y = 3

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

6x + 6y = −3——(1)

−6x−6y = 3——-(2)

After adding  Equation (1) and  Equation (2), we get

​⇒  ​6x−6x + 6y−6y = −3 + 3

​⇒  0 = 0

This is true, so the given system of equations has infinitely many solutions.

The given system of equations has infinitely many solutions.

 

Page 150  Exercise 6 Answer

Given: A system of equations is given to us. It is  2x−5y =−3, 3x + 5y = 8

To find: We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

​2x−5y=−3——(1)

3x+5y=8——–(2)

After adding  Equation (1) and Equation (2),  we get

​⇒  ​2x + 3x−5y + 5y =−3+8

​⇒  5x = 5

​⇒  x = 1

Putting x = 1 in  Equation (1)  we get

​​2x−5y=−3

⇒ 2×1−5y = −3

⇒  2 + 3 = 5y

⇒  5y = 5

⇒  y = 1

The solution of the given system of equations 2x−5y = −3, 3x + 5y = 8 is x = 1,y = 1.

 

Page 150  Exercise 7  Answer

Given: A system of equations is given to us. It is  2x+3y=1, −2x+3y=−7

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

2x+3y=1——–(1)

−2x+3y=−7—–(2)

After adding Equation  (1) and  Equation (2) we get

​⇒  2x−2x + 3y + 3y = 1−7

⇒  6y = −6

⇒  y =−1

Putting y =−1 in equation  (1), we get

2x + 3y = 1

⇒  ​2x+3×−1 = 1

⇒ 2x−3 = 1

⇒  2x = 4

⇒  x = 2


The solution of the given system of equations  2x+3y = 1,−2x + 3y = −7 is x = 2,y =−1.

 

Page 150   Exercise 8   Answer

 A system of equations is given to us. It is  4x + 3y = 17, −8x−6y = 34

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.​4x+3y=17 −8x−6y=34

The given system of equations is:

​4x+3y=17——(1)

−8x−6y=34—–(2)


Multiplying the equation (1)  by 2, we get

⇒   ​8x + 6y = 34

⇒  −8x − 6y = 34


After adding equation (1) and equation (2), we get

​⇒  8x−8x + 6y−6y = 34 + 34

⇒  0 = 68


This is not true, so the solution of the given system of equations does not exist.

The solution of the given system of equations 4x + 3y = 17,−8x−6y = 34 does not exist.

 

Page 151  Exercise 10  Answer

Given: A system of equations is given to us. It is −2x + 5y = −21, 2x−5y = 21

To find: We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables

The given system of equations is:

⇒​ −2x + 5y = −21———(1)

⇒ 2x−5y = 21————-(2)

After adding them, we get

⇒ ​−2x + 2x + 5y−5y = −21 + 21

⇒ 0 = 0


This is true, so the given system of equations has infinitely many solutions.

The given system of equations −2x + 5y = −21,2x−5y = 21 has infinitely many solutions.

 

Page 151   Exercise 11   Answer

Given: A system of equations is given to us. It is  3x−8y = 3, 8x−3y = 8

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

​3x−8y = 3———(1)

8x−3y = 8———(2)

 

After adding  equation (1) and equation  (2), we get

⇒ ​3x + 8x−8y−3y = 3 + 8

⇒  11x−11y = 11

⇒  x−y = 1


After subtracting equation (1) and equation  (2), we get

​⇒  3x−8x−8y + 3y = 3−8

⇒  −5x−5y = −5

⇒  x + y = 1

The system of equations is:

⇒   ​x−y = 1——–(3)

⇒   x + y = 1——–(4)

After adding equation (3) and equation  (4), we get

⇒​  x + x−y + y = 1 + 1

⇒  2x = 2

⇒  x = 1

Putting x=1 in equation  (1), we get

⇒  (3) ​1−y=1

⇒  0=y

The solution of the given system of equations 3x−8y = 3,8x−3y = 8 is x = 1,y = 0.

 

Page 151  Exercise 12  Answer

Given: A system of equations is given to us. It is  18x + 12y = 24, 3x + 2y = 6

To find We have to solve the given system.
We will use the elimination or the substitution method and get the value of variables.

 

The given system of equations is:

​18x + 12y = 24——–(1)

3x + 2y=6———-(2)

Multiplying  the  equation (2)  by 6, we get

⇒​  18x + 12y = 24

⇒ 18x + 12y = 36

After subtracting equation (1) and equation (2), we get

​⇒ 18x−18x + 12y−12y = 24−36

⇒ 0 = −12

This is not true, so the solution of the given system of equations does not exist.

The solution of the given system of equations 18x+12y=24,3x+2y=6 does not exist.

 

Page 151  Exercise 14  Answer

Given: The given linear equations are:​−3x−5y=8, 6x+10y=−16.

To Find The given equations are to be solved with the respective variables.
Solution: The methods of substitution and elimination are two ways any linear equation can be solved.

The given equations are​:
−3x−5y = 8  6x + 10y =−16 ,with two variables  x & y.

This could be solved by the elimination method wherein equation −3x−5y = 8 will be multiplied by 2 to get solved.
This gives us the equation −6x−10y = 16, which is added to the equation 6x + 10y =−16, which cancels all the terms giving us no solution.

 

The graph below also shows that there exists no solution as the equations plotted shows the same line:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 6

 

 

Solving the given equations −3x−5y = 8,6x + 10y = −16 we get that there exists no solution which is also proved by the graph below:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 7

 

 

Page 151  Exercise 16  Answer

Given: The linear equations given are​ 5x + 7y = 7 7x + 5y = 5.

To Find The equations are to be solved with respect to the variables x & y.
Solution: This can be solved by the estimation method and the graph.

The given equations are :
​5x + 7y = 7 7x + 5y = 5  which are solved with respect to the variables  x & y​.

Multiplying the equation 5x + 7y =7 by 7 and  7x + 5y = 5 by 5 and subtracting them we get ​35x + 49y = 49, −35x−25y = −25.

By solving that we get the values of x & y as 0,1.

 

The graph below also shows the solution as

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 8

 

 

The equations 5x + 7y = 7,7x+5y = 5 solved gives the values x = 0 & y = 1 , which is shown in the graph below:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 9

 

 

Page  151 Exercise 17  Answer

Given: The equation given is ​y=\(\frac{2}{3} x+7\), y=\(\frac{2}{3} x-5\).

To Find We need to find the values of x & y.
Solution: Solving these equations we get the values of variables shown by the graph.

The equations are rewritten as​ 3y−2x = 21 and 3y−2x =−15.
This is solved wherein we get no solution with respect to the variables.

 

The graph is shown below

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 10

The graph shows two parallel lines with no solution.

 

The solution of the​​y=\(\frac{2}{3} x+7\), y=\(\frac{2}{3} x-5\) equations are no solution also shown in the graph below:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 11

 

 

Page 151  Exercise 18  Answer

Given: The equations given are​−3x+5y=15, 9x−15y=−45

To Find The equations are to be solved and the values of variables are to be found
Solution: The equations will be solved by the elimination method.

The equation −3x + 5y = 15 is multiplied by 3.

Simplifying them we get no solution as the variables are canceled.

 

The graph of the linear equations is:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 12

And this shows that there exists no solution.

 

Solving the linear equations−3x + 5y = 15,9x−15y =−45 we get that there exists no solution which is given by the graph:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 13

 

Page 151  Exercise 19 Answer

Given: Me and my friend have 15 & 25 as savings.

To FindThe linear equations for the problem are to be found, and the situation when I and my friend will have the same amount is to be found and explained.
Solution:  The linear equations will be made and solved.

The equation will be ​15 + 5x = y 25 + 5x = y.
Solving these we will get no solution, as the variables will be canceled.

 

The graph is shown as

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 14

This will show that there will be no situation wherein there will be the same amount.

 

The linear equations are​ 15 + 5x = y, 25 + 5x = y, which is plotted on the graph is shown as:

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.4 Solving Special Systems Of Linear Equations graph 15

Which when solved gives no solution, and this shows that there will be any situation when there will be the same amount.

 

Leave a Comment