body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Pearson Algebra 1 Common Core, 2011

PA

Pearson Algebra 1 Common Core, 2011 View details

3. Solving Systems Using Elimination

Continue to next subchapter

Exercise 37 Page 383

Multiplying both equations by a constant can remove some of the fractions, which would make it easier to solve the system of equations.

(6,-4), see solution.

Practice makes perfect

Since no variable term is isolated or has a coefficient equal to 1, it is not convenient to use the Substitution Method. Therefore, we will use Elimination Method. In order to eliminate y-terms, let's multiply the first equation by 2 and the second equation by 5.

13x+ 12y=0 & (I) 12x+ 15y= 115 & (II)

(I):LHS * 2=RHS* 2

23x+y=0 & (I) 12x+ 15y=11 & (II)

(II):LHS * 5=RHS* 5

23x+y=0 & (I) 52x+y=11 & (II)

Now that we have the same coefficient for the y-variable in both equations, we can continue by subtracting Equation (I) from Equation (II).

23x+y=0 & (I) 52x+y=11 & (II)

(II): Subtract (I)

23x+y=0 52x+y-( 23x+ y)=11- 0

(II): Distribute -1

23x+y=0 52x+y- 23x-y=11- 0

(II): Simplify terms

23x+y=0 52x- 23x=11

(II): LHS * 6=RHS* 6

23x+y=0 15x-4x=66

(II): Subtract term

23x+y=0 11x=66

(II): .LHS /11.=.RHS /11.

23x+y=0 x=6

Now that we found the value of x we can substitute it into the first equation to find y.

23x+y=0 x=6

(I): x= 6

23* 6+y=0 x=6

(I): Multiply

4+y=0 x=6

(I): LHS-4=RHS-4

y=-4 x=6

The solution to the system — the point of intersection — is (6,-4).

Subchapter links

Lesson Check p.381

Practice and Problem-Solving Exercises p.382-384

Standardized Test Prep p.384

Mixed Review p.384