Exercise 8 Page 375

In the Substitution Method, we substitute an expression for a variable in an equation into the second equation. To do so, we need an isolated variable in one of the equations. There are some possible situations to consider when deciding which equation to use for our substitution.

1. Is one of the equations already solved for one of the variables? If yes, we can substitute it into the other equation.
2. Does one of the equations have a variable with a coefficient of 1? If yes, we can isolate that variable and then substitute it into the other equation.
3. Which equation has a variable with a small coefficient? We can determine that equation and use it to solve for a variable as a first step of substitution.

Now, consider the given system of linear equations. 2.5x-7y=7.5 & (I) 6x-y=1 & (II) Notice that none of the equations have an isolated variable. Also, none of them have a variable with a coefficient of 1. We can see, however, that the variable y has a coefficient of - 1 in Equation (II). We can isolate that variable in two steps. Therefore, it is a better choice to use Equation (II) in the first step of substitution to solve for a variable.

2.5x-7y=7.5 & (I) 6x-y=1 & (II)

AddEqn

(II):LHS+y=RHS+y

2.5x-7y=7.5 6x=y+1

SubEqn

(II):LHS-1=RHS-1

2.5x-7y=7.5 6x-1=y

RearrangeEqn

(II):Rearrange equation

2.5x-7y=7.5 y=6x-1

Once we obtain an isolated variable in Equation (II), we can easily substitute it into Equation (I) and solve the system of equations.