Systems of linear equations

A linear equation in two variables

Consider an equation of the form

3x+2y=0

where $x$ and $y$ are variables which stand for unknown values. So in contrast to an ordinary linear equation, e.g. $3x+2=0$ , the equation above has two unknowns. But we still call the equation $3x+2y=0$ linear, because both $x$ and $y$ are neither squared, not the root is taken, and so on. For example, all these equations are not linear:

$3x^2+2y=0$
$3x+2\sqrt{y}=0$
$3\frac{1}{y}+2y=0$
$3x^2+2y^2=0$

To solve such the equation $3x+2y=0$ means to find values for the variables $x$ and $y$ such that the left side equals the right side. It turns out that there are infinitely many solutions. For example, let's assign to $y$ an arbitrary value, say $y=4$ . So we have to find an $x$ such that

3x+2\cdot 4=0

from which follows that $x=-\frac{8}{3}$ . So a solution is

x=-\frac{8}{3}\, \text{ and }\, y=4

Or let's assign to $y$ the value $-2$ , that is, we have to find an $x$ such that

3x+2\cdot (-2)=0

from which follows $x=\frac{4}{3}$ . Thus, another solution is

x=\frac{4}{3}\, \text{ and }\, y=-2

And so on. I hope you see that in this way we can find as many different solutions as we want.

A system of linear equations

Now let's add a second equation, e.g.

\begin{array}{|lll|} 3x+2y & =&0 \\ 2x-5y & =&-1 \end{array}

We call this a system of linear equations. The two vertical lines framing the two equations are important. They indicate that we are now looking for an $x$ value and a $y$ value such that both equations are satisfied, simultaneously. Let's try one of the solutions from above, e.g.

x=-\frac{8}{3}\, \text{ and }\, y=4

Clearly, the first equation is satisfied, that is, the left side equals the right side. Hover, these values are not a solution of the second equation:

2x-5y=2\cdot \frac{8}{3}-5\cdot 4 =-\frac{44}{3}\neq -1

Solving a system of linear equations

So how can we find a solution of the system of linear equations above, if one exist at all? The method is always the same: we first solve one equation for $x$ , or for $y$ , what ever suits you. For example, let's solve the upper equation for $x$ :

\begin{array}{lll} 3x+2y & =&0& \quad | \, -2y \\ 3x & =&-2y& \quad | \, :3 \\ x & =&-\frac{2y}{3}& \end{array}

Does this help? Yes, because if we now replace the $x$ {\it in the second equation} with $-\frac{2y}{3}$ , we get a linear equation with only one variable, and this is something we can solve:

\begin{array}{lll} 2\cdot (-\frac{2y}{3}) - 5y & = & -1 \\ -\frac{4y}{3} - 5y & = & -1 \\ \frac{-4y-15y}{3} &= & -1 &\quad | \, \cdot 3 \\ -19y & = & -3& \quad | \, :19 \\ y & = & \frac{3}{19} \end{array}

Now we can use this result to find $x$ :

\begin{array}{lll} x=-\frac{2y}{3}&=&-\frac{2\cdot \frac{3}{19}}{3}\\ &=&-\dfrac{\frac{6}{19}}{\frac{3}{1}}\\ &=&-\frac{6}{19}\cdot \frac{1}{3}\\ &=&-\frac{2}{19} \end{array}

So a solution of the equation above is $x=\underline{-\frac{2}{19}}$ and $y=\underline{\frac{3}{19}}$ .

Let's verify if this is correct:

\begin{array}{lll} 3\cdot (-\frac{2}{19})+2\cdot \frac{3}{19} & =0 \quad \checkmark\\ 2\cdot (-\frac{2}{19})-5\cdot \frac{3}{19} & =-1 \quad \checkmark\\ \end{array}

Note that it is important to use both equations for finding the solutions. You cannot, for example, replace $-\frac{2y}{3}$ with the $x$ in the first equation again - this will lead to the equation $0=0$ from which we gain no further knowledge about the solution.

Exercise 1

Q1

Solve the linear system of equations. Always verify your solution!

$\begin{array}{|lll|} 6x-7y & =88 \\ x & =24 \end{array}$
$\begin{array}{|lll|} x+3y & =3 \\ y & =3x \end{array}$
$\begin{array}{|lll|} 5x-7y+9 & =0 \\ x+y & =0 \end{array}$
$\begin{array}{|lll|} 15x+4y & =18 \\ 6x+y & =0 \end{array}$
$\begin{array}{|lll|} 3x+4y & =16 \\ y & =2-x \end{array}$
$\begin{array}{|lll|} 6x+5y & =4 \\ 2x & =4y+2 \end{array}$
$\begin{array}{|lll|} 9(x-y)+24x & =100 \\ 3(x-y) & =32 \end{array}$
$\begin{array}{|lll|} 4y-19 & =y+20 \\ 5x+2y & =126 \end{array}$
$\begin{array}{|lll|} 2x-3y & =6 \\ \frac{1}{2}x+\frac{1}{3}y & =\frac{1}{6} \end{array}$
$\begin{array}{|lll|} \frac{25x-3}{12}-\frac{20y-1}{18} & =2x-y \\ \frac{x+4}{9} & =\frac{y+3}{5} \end{array}$
$\begin{array}{|lll|} 16x+17y & =99 \\ 17y+16x & =99 \end{array}$

Q2

Solve the system of equations. You first have to bring the equations into a familiar form. Always verify your solution!

$\begin{array}{|lll|} \frac{x}{x-2}-\frac{4}{y} & =1 \\ \frac{x}{2+\frac{1}{y}}= & \frac{y}{3-\frac{1}{x}} \end{array}$
$\begin{array}{|lll|} \frac{1}{y+1}+\frac{1}{y-1} & =\frac{10-x}{y^2-1} \\ \frac{1}{x-2}+ \frac{3}{y-4}& =\frac{5}{2-x} \end{array}$
$\begin{array}{|lll|} x^2-y^2 & =1000 \\ x-y & =10 \end{array}$

Note: the first equation is not a linear equation, try to solve it anyway using the same method.

Q3

Translate the following word problems into a system of linear equations, and solve it. Always start by figuring out what the unknown variables are, usually denoted by $x$ and $y$ .

The admission fee at a small fair is 1.50 sFr for children and 4.00 sFr for adults. On a certain day, 2200 people enter the fair and 5050 sFr is collected. How many children and how many adults attended?
The sum of the digits ("Ziffern") of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number. Hint: A number, e.g. $82$ can be written as $8\cdot 10+2$ .
A woman owns 21 pets. Each of her pets is either a cat or a bird. If the pets have a total of 76 legs, and assuming that none of the bird's legs are protruding from any of the cats' jaws, how many cats and how many birds does the woman own?
Two types of coffee beans are mixed. If you take from the cheap coffee beans twice as much as from the expensive one, a kilogram of the mixture costs 17.50 sFR. If you take from the expensive coffee beans twice as much as from the cheap one, a kilogram of the mixture costs 18.50 sFR. Determine the cost per kilogram of the cheap coffee and of the expensive coffee.
A passenger jet took three hours to fly 1800 miles in the direction of the jet stream. The return trip against the jet stream took four hours. What was the jet's speed in still air and the jet stream's speed?

Solution

A1

$x=24, y=8$
$x=0.3, y=0.9$
$x=-0.75, y=0.75$
$x=-2, y=12$
$x=-8, y=10$
$x=\frac{13}{17}, y=-\frac{2}{17}$
$x=\frac{1}{6}, y=-\frac{21}{2}$
$x=20, y=13$
$x=\frac{15}{13}, y=-\frac{16}{13}$
$x=5, y=2$
infinitely many solutions (both equations are identical).

A2

$x=0, y=4$ or $x=6,y=8$
infinitely many solutions
$x=55, y=45$

A3

$x$ number of children, $y$ number of adults.
$\begin{array}{|lll|} x+y & = &2200\\ 1.5x+4y&=&5050\\ \end{array}$
The solution is $x=1500$ and $y=700$ .
$x$ the number of tens, $y$ the number of ones. A number, e.g. $82$ can be written as
$8\cdot 10+ 2$
and the number $xy$ can be written as
$10x+y$
Thus, we have the system of equations
$\begin{array}{|lll|} x+y & = & 7\\ 10y+x &= &10x+y+27\\ \end{array}$
It follows $x=2$ and $y=5$ . So the number is 25.
$x$ number of cats (4 legs), $y$ number of birds (two legs).
$\begin{array}{|lll|} 4x+2y & = & 76\\ x+y &= & 21\\ \end{array}$
It follows $x=17, y=4$
$x$ price of cheap beans, $y$ price of expansive beans. In each case we buy a total of three kg, so we have
$\begin{array}{|lll|} 2x+y & = & 17.5\cdot 3\\ x+2y &= & 18.5\cdot 3\\ \end{array}$
It follows $x=16.5$ sFr/kg and y $=19.5$ sFr/kg.
$x$ speed of jet stream (miles per hour relative to ground), $y$ speed of jet without wind (miles per hour relative to ground). Total speed of plane (relative to ground) if it moves in direction of the jet stream: $y+x$ . Total speed if jet moves against the jet stream: $y-x$ .
$\begin{array}{|lll|} y+x & = & \frac{1800}{3}\\ y-x &= & \frac{1800}{4}\\ \end{array}$
It follows $x=75$ miles per hour, $y=525$ miles per hour.