Practice questions
Chapter 1 | Chapter 2 | Chapter 3 | Chapter 4 | Chapter 5 | Chapter 6
Chapter 6: Systems of equations and inequalities
Try each problem first. Then check your work with the solution.
1. Systems of equations in two variables.
(a) Solve by substitution: ${y=3x-5}$ and ${2x+y=10}$.
(b) Solve by elimination: ${4x-3y=11}$ and ${6x+5y=7}$.
(c) How many solutions does the system ${3x-6y=9}$, ${x-2y=4}$ have?
(d) A chemist needs ${20}$ L of a ${30\%}$ acid solution. A ${10\%}$ solution and a ${50\%}$ solution are on hand. How many liters of each should be mixed?
(e) Solve the system ${y=x^2-4}$ and ${y=x+2}$.
Solution:
(a)
Substitute ${y=3x-5}$ into the second equation:
$\begin{align*}2x+(3x-5)&=10\\5x&=15\\x&=3\end{align*}$
Then ${y=3(3)-5=4}$. The solution is ${(3,4)}$.
(b)
Multiply the first equation by ${3}$ and the second by ${2}$, so both ${x}$-coefficients are ${12}$:
$\begin{cases}12x-9y=33\\12x+10y=14\end{cases}$
Subtract the first from the second:
$\begin{align*}19y&=-19\\y&=-1\end{align*}$
Then ${4x+3=11}$, so ${x=2}$. The solution is ${(2,-1)}$.
(c)
Multiply the second equation by ${3}$: ${3x-6y=12}$. Subtract it from the first equation: ${0=-3}$. This is false, so the system has no solution. The lines are parallel.
(d)
Let ${x}$ be the liters of ${10\%}$ solution, and ${y}$ the liters of ${50\%}$ solution. The total volume is ${20}$ L. The acid in the mix is ${30\%}$ of ${20}$ L, which is ${6}$ L:
$\begin{cases}x+y=20\\0.1x+0.5y=6\end{cases}$
Substitute ${y=20-x}$:
$\begin{align*}0.1x+0.5(20-x)&=6\\-0.4x+10&=6\\x&=10\end{align*}$
So ${y=10}$. Mix ${10}$ L of each.
(e)
Set the two expressions for ${y}$ equal:
$\begin{align*}x^2-4&=x+2\\x^2-x-6&=0\\(x-3)(x+2)&=0\end{align*}$
So ${x=3}$ or ${x=-2}$. Then ${y=x+2}$ gives the solutions ${(3,5)}$ and ${(-2,0)}$.
2. Systems of linear equations in three variables.
Solve the system ${x+y+z=6}$, ${x-y+2z=5}$, ${2x+y-z=1}$.
Solution:
Eliminate ${x}$. Subtract the first equation from the second, and subtract ${2}$ times the first from the third:
$\begin{cases}x+y+z=6\\-2y+z=-1\\-y-3z=-11\end{cases}$
Eliminate ${y}$ from the third equation. Multiply the third by ${-2}$, and add the second:
$\begin{align*}(-2y+z)-2(-y-3z)&=-1+22\\7z&=21\\z&=3\end{align*}$
Back-substitute. From the second equation, ${-2y+3=-1}$, so ${y=2}$. From the first, ${x=6-2-3=1}$.
The solution is ${(1,2,3)}$.
3. Matrices and Gaussian elimination.
(a) Write the augmented matrix of the system ${x+2y-z=-1}$, ${2x+y+z=4}$, ${3x-y+z=8}$.
(b) Solve the system in (a) by Gauss-Jordan elimination.
(c) Each matrix below is the reduced form of a system in ${x}$, ${y}$, ${z}$. Give the solutions.
(i) $\left[\begin{array}{rrr|r}1&0&2&3\\0&1&-1&1\\0&0&0&4\end{array}\right]$
(ii) $\left[\begin{array}{rrr|r}1&0&2&3\\0&1&-1&1\\0&0&0&0\end{array}\right]$
Solution:
(a)
$\left[\begin{array}{rrr|r}1&2&-1&-1\\2&1&1&4\\3&-1&1&8\end{array}\right]$
(b)
${R_2-2R_1\to R_2}$ and ${R_3-3R_1\to R_3}$:
$\left[\begin{array}{rrr|r}1&2&-1&-1\\0&-3&3&6\\0&-7&4&11\end{array}\right]$
${-\frac{1}{3}R_2\to R_2}$:
$\left[\begin{array}{rrr|r}1&2&-1&-1\\0&1&-1&-2\\0&-7&4&11\end{array}\right]$
${R_3+7R_2\to R_3}$, then ${-\frac{1}{3}R_3\to R_3}$:
$\left[\begin{array}{rrr|r}1&2&-1&-1\\0&1&-1&-2\\0&0&1&1\end{array}\right]$
${R_2+R_3\to R_2}$ and ${R_1+R_3\to R_1}$:
$\left[\begin{array}{rrr|r}1&2&0&0\\0&1&0&-1\\0&0&1&1\end{array}\right]$
${R_1-2R_2\to R_1}$:
$\left[\begin{array}{rrr|r}1&0&0&2\\0&1&0&-1\\0&0&1&1\end{array}\right]$
The solution is ${(2,-1,1)}$.
(c)
(i) The last row says ${0=4}$. So there is no solution.
(ii) The last row is all zeros, and ${z}$ is free. Let ${z=t}$. The rows say ${x+2z=3}$ and ${y-z=1}$. So the solutions are ${(3-2t,\ 1+t,\ t)}$, for any real number ${t}$.
4. Systems of inequalities.
(a) Graph the inequality ${y<-2x+4}$.
(b) Graph the system ${x\ge 0}$, ${y\ge 0}$, ${2x+y\le 10}$, ${x+y\le 7}$, and find its vertices.
(c) Find the largest value of ${P=3x+2y}$ on the region in (b).
Solution:
(a)
Draw the boundary ${y=-2x+4}$ dashed, because the sign is ${<}$. Test ${(0,0)}$: ${0<4}$ is true. So shade the side that contains the origin.
(b)
The region is in the first quadrant, below both lines. The corners on the axes are ${(0,0)}$, ${(5,0)}$ (from ${2x+y=10}$), and ${(0,7)}$ (from ${x+y=7}$).
For the corner where the lines meet, subtract ${x+y=7}$ from ${2x+y=10}$: ${x=3}$. Then ${y=7-3=4}$. The vertex is ${(3,4)}$.
(c)
The largest value is at a vertex. Check each one:
| Corner | ${P=3x+2y}$ |
|---|---|
| ${(0,0)}$ | ${0}$ |
| ${(5,0)}$ | ${15}$ |
| ${(3,4)}$ | ${9+8=17}$ |
| ${(0,7)}$ | ${14}$ |
The largest value is ${17}$, at ${(3,4)}$.