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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 4 Page 375

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

No solution.

Practice makes perfect

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

Result of solving by substitution	Number of solutions
Values for x and y are determined.	One solution
An identity is found.	Infinitely many solutions
A contradiction is found.	No solution

This means that 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.

Consider the given equations, we need to isolate one of the variables. Let's start by isolating x in Equation I and then substituting it into Equation II.

- x+1/2y=13 & (I) x+15=1/2y & (II)

(I): LHS+x=RHS+x

1/2y=13+x & (I) x+15=1/2y & (II)

(I): LHS-13=RHS-13

1/2y-13=x & (I) x+15=1/2y & (II)

(I): Rearrange equation

x=1/2y-13 & (I) x+15=1/2y & (II)

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

x=1/2y-13 & (I) x+15=1/2y & (II)

(II): x= 1/2y-13

x=1/2y-13 & (I) 1/2y-13+15=1/2y & (II)

(II): Add terms

x=1/2y-13 & (I) 1/2y+2=1/2y & (II)

(II): LHS-1/2y=RHS-1/2y

x=1/2y-13 & (I) 2≠0 & (II)

Uh oh! Solving this system resulted in a contradiction. This means that the system has no solution.

Subchapter links

Lesson Check p.375

Practice and Problem-Solving Exercises p.375-377

Standardized Test Prep p.377

Mixed Review p.377