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Pearson Algebra 1 Common Core, 2011

PA

Pearson Algebra 1 Common Core, 2011 View details

Cumulative Standards Review

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Exercise 13 Page 659

Does either of the equations have an isolated variable in it?

A

Practice makes perfect

In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. This method consists of three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate x in the first equation.

x+2y=23 & (I) 4x-y=-7 & (II)

(I): LHS-2y=RHS-2y

x=23-2y 4x-y=-7

Now that we have isolated x, we can solve the system by substitution.

x=23-2y & (I) 4x-y=-7 & (II)

(II):x= 23-2y

x=23-2y 4( 23-2y)-y=-7

(II): Distribute 4

x=23-2y 92-8y-y=-7

(II):Subtract term

x=23-2y 92-9y=-7

(II):LHS-92=RHS-92

x=23-2y - 9y=-99

(II): .LHS /(- 9).=.RHS /(- 9).

x=23-2y y=11

Great! Now, to find the value of x, we need to substitute y=11 into either one of the equations in the given system. Let's use the first equation.

x=23-2y & (I) y=11 & (II)

(I):y= 11

x=23-2( 11) y=11

(I): Multiply

x=23-22 y=11

(I): Subtract term

x=1 y=11

The solution, or point of intersection, to this system of equations is the point (1,11). This corresponds to answer A.