
a.
To describe two ways to solve the system by elimination.
a.

Answer to Problem 6.4.1P
Addition or subtraction.
multiplication
Explanation of Solution
Given:
The two different ways of solving by elimination method is
- By addition or subtraction
- By multiplication
In addition, the coefficients of atleast 1 variable is opposite. Both the equations are added and one variable is wiped out. We substitute the value in equations after solving for another variable.
In multiplication, the equation is multiplied by a constant until the coefficient of at least 1 variable become opposite. Then the same procedure is followed as addition.
b.
To explain the incorrect.
b.

Answer to Problem 6.4.1P
The answer is followed.
Explanation of Solution
Given:
As you know, the coefficient of the variable is the one to be made opposite, not the constant. Here, the multiplication of 1st equation by 6 and second equation by 5 will make the constant equal but not the variables. Hence, it is not a useful measure to solve the system.
c.
To solve the system of equation
c.

Answer to Problem 6.4.1P
Explanation of Solution
Given:
Calculation:
Let’s assume equations be (1) and (2) respectively
To solve equations by elimination, the given equations are either added or subtracted to get an equation in only one variable.
Let’s eliminate variable x here. To eliminate a variable, the coefficients of the variable must be same in both the equations.
In equation 1, coefficient of x is -2 & in equation 2, coefficient of x is 3. To make the coefficients same, lets multiply equation 1 with
This gives us,
Now, add the equation (3) and (2)
Solving,
Now taking LCM,
Dividing and solving,
Put value of y in (1)
Solving,
So,
Dividing by -2,
Hence,
d.
d.

Explanation of Solution
Given:
Graph:
The red line represents:
The blue line represents:
Interpretation:
Here, it can be seen that the intersection point is (-2, -3). This is also the solution of the graph.
Hence, the solution is correct.
Chapter ISG Solutions
Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)
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