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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 15 Page 409

What does it mean when solving a system of equations results in an identity or a contradiction?

(-1,-2)

Practice makes perfect

To determine how many solutions this system of equations has, we will solve it by substitution. Doing so will result in one of three cases.

Result of Solving by Substitution	Number of Solutions
A value for x and y is determined.	One solution
An identity is found.	Infinitely many solutions
A contradiction is found.	No solution

This means we should solve the given system of equations, then make our conclusion based on the result.

Solve by Substitution

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.

Observing the given equations, it looks like eliminating the fractions in the first equation will also isolate y.

1/3y=7/3x+5/3

LHS * 3=RHS* 3

3(1/3y)=3(7/3x+5/3)

Distribute 3

3(1/3y)=3(7/3x)+3(5/3)

FracMultDenomToNumber

a/3* 3 = a

y=7x+5

Now that we've isolated y, we can solve the system by substitution.

y=7x+5 & (I) x-3y=5 & (II)

(II): y= 7x+5

y=7x+5 x-3( 7x+5)=5

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(II): Solve for x

(II): Distribute (-3)

y=7x+5 x-21x-15=5

(II): Subtract term

y=7x+5 -20 x-15=5

(II): LHS+15=RHS+15

y=7x+5 -20 x=20

(II): .LHS /(-20).=.RHS /(-20).

y=7x+5 x=-1

(I): x= (-1)

y=7( -1)+5 x=-1

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

y=-7+5 x=-1

(I): Add terms

y=-2 x=-1

Solving this system resulted in the values x=-1 and y=-2. Therefore, there is one solution to the system.

Subchapter links

Exercises p.408-410