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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 30 Page 376

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
A value for x and y is determined.	One solution
An identity is found, such as 2=2.	Infinitely many solutions
A contradiction is found, such as 2≠ 3.	No solution

This means that we should solve the given system of equations and 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 11y in Equation II.

17=11y+12x & (I) 12x+11y=14 & (II)

(II): LHS-12x=RHS-12x

17=11y+12x & (I) 11y=14-12x & (II)

Now that we have isolated 11y, we can solve the system by substitution.

17=11y+12x & (I) 11y=14-12x & (II)

(I): 11y= 14-12x

17=( 14-12x)+12x & (I) 11y=14-12x & (II)

(I): Remove parentheses

17=14-12x+12x & (I) 11y=14-12x & (II)

(I): Add terms

17≠14 & (I) 11y=14-12x & (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