Systems of equations in two variables

Example 1: Solve the system by substitution.

$\begin{cases}2x+y=7\\x-3y=-7\end{cases}$

Solution:

Solve the first equation for ${y}$: ${y=7-2x}$.

Substitute this into the second equation, and solve for ${x}$:

$\begin{align*}x-3(7-2x)&=-7\\x-21+6x&=-7\\7x&=14\\x&=2\end{align*}$

Then ${y=7-2(2)=3}$. The solution is ${(2,3)}$.

Check: ${2(2)+3=7}$ and ${2-3(3)=-7}$.

Example 2: Solve the system by elimination.

$\begin{cases}3x+4y=10\\5x-2y=8\end{cases}$

Solution:

The ${y}$-coefficients are ${4}$ and ${-2}$. Multiply the second equation by ${2}$, so they become ${4}$ and ${-4}$:

$\begin{cases}3x+4y=10\\10x-4y=16\end{cases}$

Add the two equations. The ${y}$-terms cancel:

$\begin{align*}13x&=26\\x&=2\end{align*}$

Put ${x=2}$ into the first equation:

$\begin{align*}6+4y&=10\\y&=1\end{align*}$

The solution is ${(2,1)}$.

Check: ${5(2)-2(1)=8}$.

Example 3: Solve the system.

$\begin{cases}2x-y=3\\-4x+2y=5\end{cases}$

Solution:

Multiply the first equation by ${2}$, and add it to the second:

$\begin{align*}4x-2y&=6\\-4x+2y&=5\\\hline 0&=11\end{align*}$

This is false, so the system has no solution. The two lines, ${y=2x-3}$ and ${y=2x+\dfrac{5}{2}}$, have the same slope. They are parallel.

Example 4: Solve the system.

$\begin{cases}x-2y=4\\-3x+6y=-12\end{cases}$

Solution:

Multiply the first equation by ${3}$, and add it to the second:

$\begin{align*}3x-6y&=12\\-3x+6y&=-12\\\hline 0&=0\end{align*}$

This is always true. The second equation is just ${-3}$ times the first, so both describe the same line. Every point on the line ${x-2y=4}$ is a solution.

To list the solutions, let ${y}$ be any number ${t}$. Then ${x=2t+4}$. The solutions are all pairs

${(2t+4,\ t)}$, where ${t}$ is any real number.

For example, ${t=0}$ gives ${(4,0)}$, and ${t=1}$ gives ${(6,1)}$.

Example 5: A school play sold ${500}$ tickets. Adult tickets cost ${\$12}$, and child tickets cost ${\$7}$. The total income was ${\$4850}$. How many tickets of each kind were sold?

Solution:

Let ${a}$ be the number of adult tickets, and ${c}$ the number of child tickets. There are two facts: the number of tickets, and the money.

$\begin{cases}a+c=500\\12a+7c=4850\end{cases}$

From the first equation, ${c=500-a}$. Substitute it into the second:

$\begin{align*}12a+7(500-a)&=4850\\12a+3500-7a&=4850\\5a&=1350\\a&=270\end{align*}$

Then ${c=500-270=230}$. So ${270}$ adult tickets and ${230}$ child tickets were sold.

Check: ${12(270)+7(230)=3240+1610=4850}$.

Example 6: A boat travels ${36}$ km down a river in ${2}$ hours. The return trip, against the current, takes ${3}$ hours. Find the speed of the boat in still water, and the speed of the current.

Solution:

Let ${b}$ be the speed of the boat in still water, and ${c}$ the speed of the current, in km/h. Going downstream, the current helps, so the speed is ${b+c}$. Going upstream, it slows the boat, so the speed is ${b-c}$.

Speed is distance divided by time:

${b+c=\dfrac{36}{2}=18}$

${b-c=\dfrac{36}{3}=12}$

Add the two equations: ${2b=30}$, so ${b=15}$. Then ${c=18-15=3}$.

The boat goes ${15}$ km/h in still water, and the current flows at ${3}$ km/h.

Example 7: Solve the system.

$\begin{cases}x^2+y^2=25\\x+y=7\end{cases}$

Solution:

Solve the linear equation for ${y}$: ${y=7-x}$. Substitute it into the first equation:

$\begin{align*}x^2+(7-x)^2&=25\\x^2+49-14x+x^2&=25\\2x^2-14x+24&=0\\x^2-7x+12&=0\\(x-3)(x-4)&=0\end{align*}$

So ${x=3}$ or ${x=4}$.

If ${x=3}$, then ${y=7-3=4}$. If ${x=4}$, then ${y=3}$. The solutions are ${(3,4)}$ and ${(4,3)}$.