Exercise 10 Page 375

We are asked to tell whether the given statement is true or false.

You cannot use substitution to solve a system that does not have a variable with a coefficient of 1 or - 1.

We can investigate this claim by creating our own arbitrary example. Consider the following system of equations. 2x=2y & (I) 3x+3y=6 & (II) Notice that it does not have a variable with a coefficient of 1 or - 1. Because we want to try to use the Substitution Method, we will start by isolating the x-variable in Equation (I).

2x=2y 3x+3y=6

DivEqn

(I): .LHS /2.=.RHS /2.

x=y 3x+3y=6

Now that the x-variable is isolated in Equation (I), we are going to substitute its equivalent expression in Equation (II).

x=y 3x+3y=6

Substitute

(II): x= y

x=y 3 y+3y=6

AddTerms

(II): Add terms

x=y 6y=6

DivEqn

(II): .LHS /6.=.RHS /6.

x=y y=1

We have found that y=1. We will find the value for x by substituting 1 for y in Equation (I).

x=y y=1

Substitute

(I): y= 1

x= 1 y=1

The solution of the system of equations, which is the point of intersection of the lines, is (1,1). We could find the solution despite the fact that the original system did not have a variable with a coefficient of 1 or - 1. We can always solve to isolate a variable before substituting. Therefore, the statement is false.