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Houghton Mifflin Harcourt Algebra 1, 2015

HM

Houghton Mifflin Harcourt Algebra 1, 2015 View details

3. Solving Linear Systems by Adding or Subtracting

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Exercise 21 Page 411

First combine all variables on one side of the second equation and than add the equations in order to eliminate the w-variable.

Length: 32 ft
Width: 15 ft

Practice makes perfect

Let's start by combining all variables on one side of the second equation.

2l+2w=94 & (I) l=2w+2 & (II)

(II):LHS-2w=RHS-2w

2l+2w=94 l-2w=2

Examining the System of Linear Equations, we see that the w-variables have opposite coefficients. This means that we can use the Elimination Method to solve for the length and width. Let's start by adding the second equation to the first and solving for the l-variable.

2l+2w=94 l-2w=2

(I):Add (II)

2l+2w+( l-2w)=94+ 2 l-2w=2

(I): Remove parentheses

2l+2w+l-2w=94+2 l-2w=2

(I):Add and subtract terms

3l=96 l-2w=2

(I):.LHS /3.=.RHS /3.

l=32 l-2w=2

Now that we have solved for the l-variable, we can substitute this value into the second equation and solve for l.

l=32 l-2w=2

(II):l= 32

l=32 32-2w=2

(II):LHS-32=RHS-32

l=32 -2w=-30

(II):.LHS /(-2).=.RHS /(-2).

l=32 w=15

We have found that length of the pool is 32 feet and width is 15 feet.

Subchapter links

Explore p.405-409

Elaborate p.409