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1Solving linear systems algebraically: substitution and linear combination

2Warmup

Try to solve this system algebraically or graphically. Remember that solving the system means finding a solution that works for all equations.

\begin{array}{rcl} 2 x + 2 y & = & 10 \\ y & = & \frac{3}{2} x \end{array}

3Misc

Collect homework

Homework for next class: Homework: Read section 3.2, p.148-151. Problems: p.152-153 1-50 odd.

4Hybrid car again

Remember the hybrid car from the test? We said it gets 40 miles/gallon and 2 miles/kw-h. Let's also estimate that a gallon costs $3 and a kw-h costs $.1. How would we find out how much of each kind of fuel to use if we wanted to go 200 miles and spend $13?

5Substitution

Logic: You can substitute equal things for each other without changing the solution. This method is especially convenient when a coefficient is 1.

5.1Example

Solve:

\begin{array}{rcl} 3 x + 4 y & = & - 4 \\ x + 2 y & = & 2 \end{array}

Solution: Solve one equation for $x$ or $y$ , making a revised equation:

$\begin{array}{lll} x + 2 y & = & 2 \\ x & = & - 2 y + 2 \end{array}$

...and substitute in the other equation and solve for the remaining variable:

\begin{array}{rcl} 3 x + 4 y & = & - 4 \\ 3 (- 2 y + 2) + 4 y & = & - 4 \\ - 6 y + 6 + 4 y & = & - 4 \\ - 2 y & = & - 10 \\ y & = & 5 \end{array}

...then plug that in and solve the revised equation for the first variable:

\begin{array}{rcl} x & = & - 2 y + 2 \\ = & - 2 \cdot 5 + 2 \\ = & - 8 \end{array}

So the solution is $(- 8, 5)$ .

Check your work! Make sure you have a solution. For example, plug in to the first equation:

\begin{array}{rcl} 3 x + 4 y & = & 3 \cdot (- 8) + 4 \cdot 5 \\ = & - 24 + 20 \\ = & - 4 ? \end{array}

Graph it to visualize.

6Linear combination

6.1Example

The simplest linear combination case is when the two equations add up to yield a new equation in only one variable:

\begin{array}{rcl} 2 x + y & = & 3 \\ - 3 x - y & = & - 5 \end{array}

You can add them to get

\begin{array}{rcl} - x & = & - 2 \\ x & = & 2 \end{array}

Then you can plug $x$ in to either original equation to get $y$ :

\begin{array}{rcl} 2 x + y & = & 3 \\ 2 \cdot 2 + y & = & 3 \\ 4 + y & = & 3 \\ y & = & - 1 \end{array}

Then check by plugging $x$ and $y$ in to the original equations. For example: $2 x + y = 2 \cdot 2 + (- 1) = 3 ?$

6.2Example

Solve:

\begin{array}{rcl} 2 x - 4 y & = & 13 \\ 4 x - 5 y & = & 8 \end{array}

Solution: Multiply both sides of one equation so, when added to the other equation, one variable goes away:

\begin{array}{rcl} - 2 (2 x - 4 y) & = & - 2 \cdot 13 \\ 4 x - 5 y & = & 8 \end{array}

The point of this is to get rid of the $x$ term, which we see when we simplify...

\begin{array}{rcl} - 4 x + 8 y & = & - 26 \\ 4 x - 5 y & = & 8 \\ － － － & － - \\ 3 y & = & - 18 \\ y & = & - 6 \end{array}

...then use substitution for the first variable in either of the first equations:

\begin{array}{rcl} 13 & = & 2 x - 4 y \\ = & 2 x - 4 \cdot (- 6) \\ = & 2 x + 24 \\ - 11 & = & 2 x \\ - \frac{11}{2} & = & x \end{array}

So the solution is $($ $- \frac{11}{2}, - 6)$ .

Check: See if that really is a solution by plugging in to the equations. For example:

\begin{array}{rcl} 4 x - 5 y & = & 4 (- \frac{11}{2}) - 5 (- 6) \\ = & - 22 + 30 \\ = & 8 ? \end{array}

6.3Example: the hybrid car

\begin{array}{rcl} 40 x + 2 y & = & 200 \\ 3 x + . 1 y & = & 13 \end{array}

Let's multiply both sides of the 2nd equation by $- 20$ to get the $y$ terms to disappear:

\begin{array}{rcl} 40 x + 2 y & = & 200 \\ - 60 x + (- 2) y & = & - 260 then,
          adding: \\ － － － － － & － － \\ - 20 x & = & - 60 \\ x & = & 3 \end{array}

So we need 3 gallons of gas. Then we can substitute to find how much electric charge we need:

\begin{array}{rcl} 40 x + 2 y & = & 200 \\ 40 \cdot 3 + 2 y & = & 200 \\ 120 + 2 y & = & 200 \\ 2 y & = & 80 \\ y & = & 40 \end{array}

So we need 40 kw-h of electricity. We can check: $3 x + . 1 y = 3 \cdot 3 + . 1 \cdot 40 = 9 + 4 = 13 ?$ Also, $40 x + 2 y = 40 \cdot 3 + 2 \cdot 40 = 120 + 80 = 200 ?$ .

6.4Example

You can multiply each equation by a number and add. For example:

Solve:

\begin{array}{rcl} 7 x - 12 y & = & - 22 \\ - 5 x + 8 y & = & 14 \end{array}

We can make the $y$ term coefficients opposites this way:

\begin{array}{rcl} 2 (7 x - 12 y) & = & 2 (- 22) \\ 3 (- 5 x + 8 y) & = & 3 (14) \end{array}

Using the distributive law:

\begin{array}{rcl} 14 x - 24 y & = & - 44 \\ - 15 x + 24 y & = & 42 \\ － － － － & － - \\ - 1 x & = & - 2 \\ x & = & 2 \end{array}

Then we can solve for $y$ using one of the other equations. Of course we could have just multiplied the 2nd equation by $\frac{3}{2}$ , but that would have involved fractions.

7No solution

Sometimes a system has no solution:

\begin{array}{rcl} x - 2 y & = & 3 \\ 2 x - 4 y & = & 7 \end{array}

Try the linear combination method; multiply both sides of the 1st equation by $- 2$ :

\begin{array}{rcl} - 2 x + 4 y & = & 3 \\ 2 x - 4 y & = & 7 \\ － － － - & － \\ 0 & = & 10 \end{array}

This obviously has no solution (no $(x, y)$ pair can make it true).

If instead we had ended up with something like

3 = 3

we would know we had infinitely many solutions.