A one pound mix consisting of 30% cashews and 70% pistachios sells for $6.25. A one pound mix consisting of 80% cashews and 20% pistachios sells for $7.50. How would a mix consisting of 50% of each type of nut sell for?

### Solving Systems by Multiplying Both Equations

In the linear systems in this lesson, we will need to multiply both equations by a constant in order to have opposite coefficients of one of the variables. In order to determine what numbers to multiply by, we will be finding the least common multiple of the given coefficients. Recall that the least common multiple of two numbers is the smallest number which is divisible by both of the given numbers. For example, 12 is the least common multiple of 4 and 6 because it is the smallest number divisible by both 4 and 6.

Let's solve the following systems using linear combinations.

\begin{align*}2x-5y &= 15\\ 3x+7y &= 8\end{align*}

In this problem we cannot simply multiply one equation by a constant to get opposite coefficients for one of the variables. Here we will need to identify the least common multiple of the coefficients of one of the variables and use this value to determine what to multiple each equation by. If we look at the coefficients of \begin{align*}x\end{align*}, 2 and 3, the least common multiple of these numbers is 6. So, we want to have the coefficients of \begin{align*}x\end{align*} be 6 and -6 so that they are opposites and will cancel when we add the two equations together. In order to get coefficients of 6 and -6 we can multiply the first equation by 3 and the second equation by -2 (it doesn’t matter which one we make negative.)

\begin{align*}&\quad \ 3(2x-5y=15) \quad \Rightarrow \qquad \ \cancel{6x}-15y = 45\\ & -2(3x+7y=8) \qquad \quad \ + \ \underline{- \cancel{6x}-14y = -16 \; \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad -29y=29\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad y=-1\end{align*}

Now find \begin{align*}x\end{align*}:

\begin{align*}2(x)-5(-1) &= 15\\ 2x+5 &= 15\\ 2x &= 10 \\ x &= 5 \end{align*}

Solution: (5, -1)

Check your answer:

\begin{align*}2(5)-5(-1) &= 10 + 5 =15\\ 3(5)+7(-1) &= 15-7=8 \end{align*}

\begin{align*}^*\end{align*} This problem could also be solved by eliminating the \begin{align*}y\end{align*} variables first. To do this, find the least common multiple of the coefficients of \begin{align*}y\end{align*}, 5 and 7. The least common multiple is 35, so we would multiply the first equation by 7 and the second equation by 5. Since one of them is already negative, we don’t have to multiply by a negative.

\begin{align*}7x+20y &= -9\\ -2x-3y &= 8\end{align*}

The first step is to decide which variable to eliminate. Either one can be eliminated but sometimes it is helpful to look at what we need to multiply by to eliminate each one and determine which is easier to eliminate. In general, it is easier to work with smaller numbers so in this case it makes sense to eliminate \begin{align*}x\end{align*} first. The Least Common Multiple (LCM) of 7 and 2 is 14. To get 14 as the coefficient of each term, we need to multiple the first equation by 2 and the second equation by 7:

\begin{align*}& 2(7x+20y=-9) \quad \Rightarrow \quad \ \ \cancel{14x}+40y= -18 \\ & 7(-2x-3y=8) \qquad \qquad \ \underline{- \cancel{14x}-21y=56 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 19y = 38\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ y=2\end{align*}

Now find \begin{align*}x\end{align*}:

\begin{align*}-2x-3(2) & = 8\\ -2x-6 &= 8\\ -2x &= 14\\ x &= -7\end{align*}

Solution: (-7, 2)

Check your answer:

\begin{align*}7(-7)+20(2) &= -49 + 40 = -9\\ -2(-7)-3(2) & =14-6=8 \end{align*}

\begin{align*}14x-6y &= -3\\ 16x-9y &= -7\end{align*}

This time, we will eliminate \begin{align*}y\end{align*}. We need to find the LCM of 6 and 9. The LCM is 18, so we will multiply the first equation by 3 and the second equation by -2. Again, it doesn’t matter which equation we multiply by a negative value.

\begin{align*}& \quad 3(14x-6y=-3) \quad \Rightarrow \quad \ + 42x - \cancel{18y} = -9\\ & -2(16x-9y=-7) \qquad \quad \ \ \underline{-32x+ \cancel{18y} = 14 \; \; \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 10x = 5\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x=2\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*} 14 \left(\frac{1}{2} \right)-6y &= -3\\ 7-6y &= -3\\ -6y &= -10\\ y & = \frac{10}{6} = \frac{5}{3}\end{align*}

Solution: \begin{align*}\left(\frac{1}{2}, \frac{5}{3}\right)\end{align*}

Check your answer:

\begin{align*}14 \left(\frac{1}{2}\right) - 6 \left(\frac{5}{3}\right) &= 7-10=-3\\ 16 \left(\frac{1}{2}\right) - 9 \left(\frac{5}{3}\right) &= 8-15=-7 \end{align*}

### Examples

#### Example 1

Earlier, you were asked to find how much a mix consisting of 50% of each type of nut would sell for.

Write a system of linear equations to represent the given information. Let \begin{align*}x=\end{align*} the cost of the cashews per pound and let \begin{align*}y=\end{align*} the cost of the pistachios per pound. Now we can write two equations to represent the two different mixes of nuts:

\begin{align*}0.3x+0.7y &= 6.25\\ 0.8x+0.2y &= 7.50\end{align*}

Solve this system to determine the cost of each type of nut per pound. If we eliminate \begin{align*}y\end{align*}, we will need to multiple the first equation by 2 and the second equation by -7:

\begin{align*}& \quad 2(0.3x-0.7y=6.25) \quad \Rightarrow \quad +0.6x - \cancel{1.4y} = 12.5\\ & -7(0.8x+0.2y=7.50) \qquad \quad \ \underline{-5.6x+ \cancel{1.4y} = -52.5 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -5x = -40\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x=8\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*}0.3(8)+0.7y &= 6.25\\ 2.4+0.7y &= 6.25\\ 0.7y &= 3.85\\ y &= 5.5\end{align*}

So, we have determined that the cost of the cashews is $8 per pound and the cost of the pistachios is $5.50 per pound. Now we can determine the cost of the 50% mix as follows:

\begin{align*}0.5(8.00)+0.5(5.50)=4.00+2.75=6.75\end{align*} So, the new mix is $6.75 per pound.

**Solve the following systems using linear combinations.**

#### Example 2

\begin{align*}6x+5y &= 3\\
-4x-2y &= -14\end{align*}

We can eliminate either variable here. To eliminate \begin{align*}x\end{align*}, we can multiple the first equations by 2 and the second equation by 3 to get 12 - the LCM of 6 and 4.

\begin{align*}& \ \ 2(6x+5y=3) \quad \ \Rightarrow \quad \ +\cancel{12x} +10y = 6\\ & 3(-4x-2y=-14) \qquad \quad \quad \underline{ - \cancel{12x}-6y = -42 \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 4y = -36\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ y=-9\end{align*}

Now find \begin{align*}x\end{align*}:

\begin{align*}6x+5(-9) &= 3\\ 6x-45 &= 3\\ 6x &= 48\\ x &= 8\end{align*}

Solution: (8, -9)

#### Example 3

\begin{align*}9x-7y &= -19\\ 5x-3y &= -15\end{align*}

Again we can eliminate either variable. To eliminate \begin{align*}y\end{align*}, we can multiply the first equation by 3 and the second equation by -7:

\begin{align*} & \quad \ 3(9x-7y=-19) \quad \Rightarrow \ \ \quad +27x - \cancel{21y} = -57\\ & -7(5x-3y=-15) \qquad \quad + \ \underline{-35x+\cancel{21y} = 105 \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ -8x = 48\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x=-6\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*}5(-6)-3y &= -15\\ -30-3y &= -15\\ -3y &= 15\\ y &= -5\end{align*}

Solution: (-6, -5)

#### Example 4

\begin{align*}15x-21y &= -63\\
7y &= 5x+21\end{align*}

To start this one we need to get the second equation in standard form. The resulting system will be:

\begin{align*}15x-21y &= -63\\ -5x+7y &= 21\end{align*}

This time we just need to multiply the second equation by 3 to eliminate \begin{align*}x\end{align*}:

\begin{align*}& \ \ 15x-21y=-63 \quad \Rightarrow \quad + \cancel{15x} - 21y = -63\\ & 3(-5x+7y=21) \qquad \ + \ \underline{- \cancel{15x}+21y = -63 \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 0y = 0\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0=0 \end{align*}

Solution: There are infinite solutions.

### Review

Solve the systems using linear combinations.

- \begin{align*}17x-5y &= 4\\ 2x+7y &= 46\end{align*}
- \begin{align*}9x+2y &= -13\\ 11x+5y &= 2\end{align*}
- \begin{align*}3x+4y &= -16\\ 5x+5y &= -5\end{align*}
- \begin{align*}2x-8y &= -8\\ 3x+7y &= 45\end{align*}
- \begin{align*}5x-10y &= 60\\ 6x+3y &= -33\end{align*}
- \begin{align*}3x+10y &= -50\\ -5x-7y &= 6\end{align*}
- \begin{align*}11x+6y &= 30\\ 13x-5y &= -25\end{align*}
- \begin{align*}15x+2y &= 23\\ 18x-9y &= -18\end{align*}
- \begin{align*}12x+8y &= 64\\ 17x-12y &= 9\end{align*}
- \begin{align*}11x-3y &= 12\\ 33x-36 &= 9y\end{align*}
- \begin{align*}4x+3y &= 0\\ 6x-13y &= 35\end{align*}
- \begin{align*}18x+2y &= -2\\ -12x-3y &= -7\end{align*}
- \begin{align*}-6x+11y &= -109\\ 8x-15y &= 149\end{align*}
- \begin{align*}8x &= -5y-1\\ -32x+20y &= 8\end{align*}
- \begin{align*}10x-16y &= -12\\ -15x+14y &= -27\end{align*}

Set up and solve a system of equations to answer the following questions.

- A mix of 35% almonds and 65% peanuts sells for $5.70. A mix of 75% almonds and 25% peanuts sells for $6.50. How much should a mix of 60% almonds and 40% peanuts sell for?
- The Robinson family pays $19.75 at the movie theater for 3 medium popcorns and 4 medium drinks. The Jamison family pays $33.50 at the same theater for 5 medium popcorns and 7 medium drinks. How much would it cost for a couple to get 2 medium drinks and 2 medium popcorns?
- A cell phone company charges extra when users exceed their included call time and text message limits. One user paid $3.24 extra having talked for 240 extra minutes and sending 12 additional texts. A second user talked for 120 extra minutes and sent 150 additional texts and was charged $4.50 above the regular fee. How much extra would a user be charged for talking 140 extra minutes and sending 200 additional texts?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 3.8.