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Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
Chapter Closure
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Exercise 127 Page 162

a In this system of equations, at least one of the variables has a coefficient of Therefore, we will approach its solution with the Substitution Method.

When solving a system of equations using this method, 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.
b Since neither equation has a variable with a coefficient of we will use the Elimination Method to solve this system. To do this, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation.
Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (II) by the -terms will have opposite coefficients.
c In this system of equations, at least one of the variables has a coefficient of Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using this method, 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.
For this exercise, is already isolated in one equation, so we can skip straight to solving!
d In this system of equations, at least one of the variables has a coefficient of Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using this method, 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.
Observing the given equations, it looks like it will be simplest to isolate in the first equation.
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Exercises 129 (Page 162)
Exercises 130 (Page 163)
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Exercises 133 (Page 163)
Exercises 134 (Page 164)
Exercises 135 (Page 164)
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