Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
2. Solving Systems Using Substitution
Continue to next subchapter

Exercise 31 Page 376

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

One 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.

  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. 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 3x in Equation I.
1.5x+2y=11 & (I) 3x+6y=22 & (II)
3x+4y=22 & (I) 3x+6y=22 & (II)
3x=22-4y & (I) 3x+6y=22 & (II)
Now that we've isolated 3x, we can solve the system by substitution.
3x=22-4y & (I) 3x+6y=22 & (II)
3x=22-4y & (I) 22-4y+6y=22 & (II)
3x=22-4y & (I) 22+2y=22 & (II)
3x=22-4y & (I) 2y=0 & (II)
3x=22-4y & (I) y=0 & (II)
Now, to find the value of x, we need to substitute y=0 into either one of the equations in the given system. Let's use the first equation.
3x=22-4y & (I) y=0 & (II)
3x=22-4( 0) & (I) y=0 & (II)
3x=22 & (I) y=0 & (II)
x= 223 & (I) y=0 & (II)
Solving this system resulted in the values x= 223 and y=0. This is the one solution to this system.