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

**Page 152 Exercise 1 Answer**

**Given:** The equation 2x−1\(=\frac{-1}{2} x\)+4 is given.

To Find The linear equations are to be found by both sides of the equation.

**Solution:** The linear equation will be by putting the equations equal to some other variable.

The given equation is 2x−1\(=\frac{-1}{2} x\)+4 which has the variable x.

The left side of the equation which is made a linear equation is y=2x−1

The right side of the equation, made a linear equation is y\(=\frac{-1}{2} x\)+4

The linear equations on the left and right sides are y=2x−1 , y\(=\frac{-1}{2} x\)+4 respectivily .

**Given:** The equation given is 2x−1 \(=\frac{-1}{2} x\)+4

To Find The graphs are to be plotted of the linear equations with the intersection point of the graphs and finding that whether the x of the intersection point is the same as the solution of the given equation.

**Solution:** The graphs will be plotted, solved, and observed to answer the questions.

**The linear equation y=2x−1 have the graph:**

**The linear equation of the right side is y\(=\frac{-1}{2} x\)+4 which is plotted as:**

**The graph plotted of the linear equations and the given equation is:**

The graph shows that the point of intersection is(2,3), wherein the x of the intersection point is the same as the solution of the given equation, i.e.,2

**The graph of the two linear equations are:**

The intersection point is(2,3), and the x of the intersection point is the same as the solution of the given equation that is 2.

**Given: **The equation is given and the linear equations are plotted on the graph to solve that equation.

To Find The reason as to why the method of the graph is better is to be answered.

**Solution:** The graph will be observed to explain the reason that the graphical method is the better one.

The equation given is 2x−1\(=\frac{-1}{2} x\)+4

The equation is plotted on the graph which is easy to be observed and explained.

The intersection points are also easy to be plotted and observed.

The graphical methods work in this case as the linear equations are properly plotted on the graph and the intersection point and the solution of the equation both can be observed and answered easily as the points are clear in the graphs.

**Page 153 Exercise 4 Answer**

We have to compare algebraic method and the graphical method for solving a linear equation with variables on both sides.

The algebraic method has advantage to be more direct as it does not involve building a system of equations, drawing graphs, and checking if the graphical solution fits the original equation. Its disadvantages might be that it involves more computation.

The advantage of graphical method is that it involves less computation. Its disadvantage is that we have to perform graphing and check the results.

Algebraic method has more computation while graphical method has less computation but it involves drawing graph and checking result.

**Page 155 Exercise 2 Answer**

**Given:** 3x=x−4

Graph the system of linear equation.

Graph the equation and find the value of x for the solution.

It can be written as two equations as shown below.

⇒y = 3x

⇒ y = x−4

The graph for these two equations is sketched on the same coordinate plane and the point of intersection is determined.

The graph for the system is sketched using a graphing calculator as shown below.

The value of at the point of intersection is −2. Therefore, the solution to the given system is determined to be x = −2

Checking the solution x = −2 for the given system.

3x = x−4

⇒ 3×−2 = −2−4

⇒ −6 = −6

The statement is true, therefore the solution is verified.

**The solution to the system of equations 3x=x−4 is determined by graphing to be x= −2. The graph is sketched below.**

**Page 155 Exercise 3 Answer**

**Given:** 4x + 1 = −2x−5

Graph the system of linear equation.

Graph the equation and find the value of x for solution.

It can be written as two equations as shown below.

⇒ y = 4x + 1

⇒ y = −2x−5

The graph for these two equations is sketched on the same coordinate plane and the point of intersection is determined.

The graph for the system is sketched using a graphing calculator as shown below.

The value of at the point of intersection is −1. Therefore, the solution to the given system is determined to be x = −1.

Checking the solution x = −1 for the given system.

4x + 1 = −2x−5

⇒ 4(−1)+1 = −2(−1)−5

⇒ −3 = 2−5

⇒ −3 =−3.

The statement is true, therefore the solution is verified.

**The solution to the system of equations 4x + 1 = −2x−5 is determined by graphing to be x = −1. The graph is sketched below.**

**Page 155 Exercise 5 Answer**

**Given:** −3x−5=6−3x

Graph the system of linear equation.

Graph the equation and find the value x of for solution.

It can be written as two equations as shown below.

⇒ y = −3x−5

⇒ y = 6−3x

The graph for these two equations is sketched on the same coordinate plane and the point of intersection is determined.

The graph for the system is sketched using a graphing calculator as shown below.

There is no point of intersection. Therefore, there is no solution to the given system.

**The solution to the system of equations 3x-5=6-3x doesn’t exist. The graph is sketched below.**

**Page 155 Exercise 6 Answer**

**Given:** 7x−14 = −7(2−x)

Graph the system of linear equation.

Graph the equation and find the value of x for solution.

It can be written as two equations as shown below.

⇒ y = 7x−14

⇒ y = −14 + 7x

The graph for these two equations is sketched on the same coordinate plane and the point of intersection is determined.

The graph for the system is sketched using a graphing calculator as shown below.

The two lines coincide. Therefore, there are infinite number of solutions to the given system.

**The solution to the system of equations 7x−14 = −7(2−x) are infinite in number. The graph is sketched below.**

**Page 156 Exercise 7 Answer**

**Given:** |3x|=|2x + 10|

Graph the system of linear equation.

Graph the equation and find the value of x for solution.

It can be written as two equations as shown below.

⇒ y = 3x

⇒ y = 2x + 10

The graph for these two equations is sketched on the same coordinate plane and the point of intersection is determined.

The graph for the system is sketched using a graphing calculator as shown below.

The value of at the point of intersection is 10. Therefore, the solution to the given system is determined to be x=10.

Checking the solution x = 10 m for the given system.

|3x| = |2x + 10|

⇒ |30| |3×10| = |2×10 + 10|

⇒ |30| = |20 + 10|

⇒ |30|=|30|

⇒ 30=30.

The statement is true, therefore the solution is verified.

**The solution to the system of equations |3x| = |2x + 10| is determined by graphing to be x = 10. The graph is sketched below.**

**Page 156 Exercise 8 Answer**

**Given:** The linear equation ∣x−1∣ = ∣x + 3∣.

We asked to find the solution of the above linear equation by graphing.

To solve the above we will consider the system of equations and graph them.

Let y = ∣x−1| and y = |x + 3| be two linear equations.

Graph of y = |x−1| and y=|x+3| is given by

From the graph the point of intersection for the system y = ∣x−1|, y = |x+3| is(−1,2).

Since |x−1| = |x+3|

Here, the solution of the above equation is the x

value of the intersection point of the system y = |x−1|, y = |x+3| in the above graph.

Since(−1,2) is the point of intersection for the system y = |x−1|, y = |x+3|.

Therefore from the graph the value of x for the equation |x−1| = |x+3| is −1.

Since |x−1|=|x + 3|.

Put x=−1 in the above equation, and we get

|−1−1|=|−1 + 3|

⇒ |−2|=|2|

⇒ 2=2

Since|−a|= an aϵℜ.

Hence the equation |x−1|=|x+3| is true for x=−1.

Therefore −1 is the solution of the equation |x−1|=|x+3|.

**-1 is the solution of the equation |x−1|=|x+3|:**

**Page 156 Exercise 9 Answer**

**Given:** The linear equation |x + 4| = |2−x|.

We asked to find the solution of the above linear equation by graphing. To solve the above we will consider the system of equations and graph them.

Let y = |x + 4| and y =|2−x| be two linear equations.

Graph of y = |x + 4| and y = |2−x| is given by

From the graph the point of intersection for the system y = |x + 4|, y = |2−x| is(−1,3).

Since∣x+4∣=∣2−x∣.

Here, the solution of the above equation is the x value of the intersection point of the system y=∣x+4∣

y=∣2−x∣ in the above graph.

Since(−1,3) is the point of intersection for the system y=|x+4| ,y=∣2−x∣.

Therefore from the graph the value of x for the equation ∣x+4∣=∣2−x∣ is−1.

Since |x+4|=|2−x|

Put x=−1 in the above equation, and we get

|−1+4|=|2−(−1)|

⇒ |3| =|2+1|

⇒ |3| = |3|

⇒ 3 = 3

Hence the equation ∣x+4∣=∣2−x∣ is true for x = −1.

Therefore−1 is the solution for the equation |x+4| = |2−x|.

**-1 is the solution for the equation ∣x+4∣=∣2−x∣. And the graph is:**