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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Systems Using Substitution

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Exercise 15 Page 375

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

(3,0)

Practice makes perfect

When solving a system of equations using substitution, there are 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.

For this exercise, y is already isolated in one equation, so we can skip straight to solving!

y=-2x+6 & (I) 3y-x+3=0 & (II)

(II):y= -2x+6

y=-2x+6 3( -2x+6)-x+3=0

(II):Distribute 3

y=-2x+6 -6x+18-x+3=0

(II):Add and subtract terms

y=-2x+6 -7x+21=0

(II):LHS-21=RHS-21

y=-2x+6 -7x=-21

(II): .LHS /-7.=.RHS /-7.

y=-2x+6 x=3

Now, to find the value of y, we need to substitute x=3 into either one of the equations in the given system. Let's use the first equation.

y=-2x+6 x=3

(I):x= 3

y=-2( 3)+6 x=3

(I): (- a)b = - ab

y=-6+6 x=3

(I): Add terms

y=0 x=5

The solution, or point of intersection, to this system of equations is the point (3,0).

Checking Our Answer

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

y=-2x+6 & (I) 3y-x+3=0 & (II)

(I), (II): x= 3, y= 0

0? =-2( 3)+6 3( 0)- 3+3? =0

(I), (II): Multiply

0? =-6+6 -3+3? =0

(I), (II): Add terms

0=0 0=0

Because both equations are true statements, we know that our solution is correct.

Subchapter links

Lesson Check p.375

Practice and Problem-Solving Exercises p.375-377

Standardized Test Prep p.377

Mixed Review p.377