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Section 3.4: Solving Systems by Substitution Worksheet 3.4

We've talked about several ways of solving systems: graphically, with a table, and numerically (with a computer).

Question: Are there other ways?

Answer: Of course! :D

Example: Let's solve a system (which is by now an an old, familiar friend) using a new ALGERBRAIC method. $$\left\{\begin{array}{l}-3x-5y=-1\\x+2y=1\\\end{array}\right\}$$









The Method of Substitution: An Algebraic Method

Step 0: If there are fractions, clear them.

Step 1: Use your knowledge of algebra to solve for one variable (either $x$ or $y$ will work; you'll get either $x=$STUFF or $y=$STUFF).

Step 2: Make a substitution: take your isolated variable and put all the other STUFF into the other equation.

Step 3: After Step 2, one of the variables will magically disappear, that is, you have an equation in 1 variable which you will solve. (Either $x=$NUMBER or $y=$NUMBER).

Step 4: Plug the NUMBER you get into any of the two original equations. This will give you another equation in the other variable. Solve this equation to get the other number.

Step 5: Check your work!











A Full-Tilt Example

Solve the following system by substitution.

$$\left\{\begin{array}{l}-\frac{x}{8}+\frac{y}{8}=\frac{1}{16}\\-\frac{x}{2}-\frac{y}{4}=-\frac{1}{4}\\\end{array}\right\}$$











Question: But what happens if there are infinitely many or no solutions?

Answer: You get either a true statement (inifitely many) or a false statement (no solution).









Example: Solve the following system. $$\left\{\begin{array}{l}-2x+6y=-2\\-x+3y=-1\\\end{array}\right\}$$