This is a paragraph.

Linear systems via elimination


In order to solve a linear system in two variables using elimination method, follow these steps:

  • Decide on which of the two variables to eliminate (this choice can be made easy if one of the two variables has a coefficient of \pm 1);
  • Ensure that the chosen variable has the same coefficients in both equations (if not, carry out a preliminary multiplication to force equality);
  • Add/subtract the equations to eliminate the variable with the same coefficient;
  • Solve the resulting one-variable linear equation and substitute back your answer to get the other variable;
  • Check that your solution satisfies the original equations.

Example 1

Solve the linear system:

(1)   \begin{equation*} x+y=7 \end{equation*}

(2)   \begin{equation*} x-y=3 \end{equation*}

Either x or y can be easily eliminated because of the numerical value of their coefficients. Suppose we choose to eliminate y, then we have to add the two equations (because then y+-y=0). So:

    \begin{equation*} 2x=10 \implies \frac{2x}{2}=\frac{10}{2} \implies x=5 \end{equation}

Next, we substitute this value of x into any of the original equations, say the first equation:

    \begin{equation*} 5+y=7 \implies y=7-5=2 \end{equation}


    \[ \boxed{x=5, y=2} \]

is the solution. Finally, we check that the solution satisfies the original equations:

    \begin{equation*} 5+2=7,\quad \textrm{and} \quad 5-2=3. \end{equation}

That’s it! Easy; very easy.

Example 2

Solve the linear system

(3)   \begin{equation*} 3x+2y = 7 \end{equation*}

(4)   \begin{equation*} 5x+13y = 31 \end{equation*}

Here, neither of the two variables has the same coefficients, so we’ll need a preliminary multiplication, whichever variable is chosen for elimination. We don’t know why we often like to eliminate y; we’re doing it again. Multiply equation (3) by 13, and then multiply equation (4) by 2 to make the coefficient of y equal to 26 in both equations:

    \begin{equation*} 39x + 26y = 91 \end{equation}

    \begin{equation*} 10x + 26y = 62 \end{equation}

Then subtract :

    \begin{equation*} 29x = 29 \end{equation}

\implies \frac{29x}{29} = \frac{29}{29} \implies x=1.

Substitute x=1 into equation (3) and solve for y:

    \begin{equation*} 3(1) + 2y = 7 \implies 3 + 2y = 7 \implies 2y = 4 \implies y=2. \end{equation}

Finally, we check that the solution (x=1,y=2) satisfies the original equations:

    \begin{equation*} 3(1) + 2(2) = 7; \quad \textrm{and}\quad 5(1) + 13(2) = 31. \end{equation}


Example 3

Solve the linear system:

(5)   \begin{equation*} 2x = 15 - 3y \end{equation*}

(6)   \begin{equation*} x + y = 5 \end{equation*}

The variable y is on the right side in equation (5), so we first re-arrange the equation:

    \begin{equation*} 2x+3y=15 \end{equation}

Multiply equation (6) by 3 (since we want to eliminate y):

    \begin{equation*} 3x+3y=15 \end{equation}



Substitute x=0 into any of the previous equations and obtain y=5. We leave the verification of the solution to you.

Don’t always expect the solution to a linear system to be integers. Sometimes, it may not be.

Example 4

Solve the linear system

(7)   \begin{equation*} 2x+y = 2 \end{equation*}

(8)   \begin{equation*} 4x-y = 1 \end{equation*}

    \begin{equation*} \textrm{\textbf{Add:}}~6x = 3, \\ \implies x=\frac{3}{6}=\frac{1}{2} \end{equation}

Substitute x=\frac{1}{2} into any of the original equations (7) and (8) to get y=1. Finally, verify the solutions.

Example 5 (an inconsistent system)

Solve the linear system

(9)   \begin{equation*} x-y = 3 \end{equation*}

(10)   \begin{equation*} 5x-5y = 10 \end{equation*}

Divide both sides of equation (10) by 5 to obtain x-y=2. So, the refactored equations are:

    \begin{equation*} x-y=3 \end{equation}

    \begin{equation*} x-y=2 \end{equation}

Subtract: 0=1, which is impossible. So there’s no solution to the original linear system; the equations simply represent two straight lines with the same slope but different y-intercepts (that is, parallel lines).

Example 6 (a system with infinitely many solutions)

Solve the linear system

(11)   \begin{equation*} 3x-2y=5 \end{equation*}

(12)   \begin{equation*} 6x-4y=10 \end{equation*}

Observe that equation (12) =2\times equation (11), so the two equations actually represent the same straight line (that is, they are coincident). Such a system possesses infinitely many solutions.

Example 7 (a system with fractional coefficients)

Solve the linear system

(13)   \begin{equation*} \frac{x}{2} + y = 7 \end{equation*}

(14)   \begin{equation*} \frac{3x}{4} - \frac{2y}{5} = 1 \end{equation*}

We first clear fractions from each equation by multiplying by the appropriate lowest common denominator or LCD (for equation (13) it is 2; while the LCD for equation (14) is 20). This way we obtain the refactored equations:

    \begin{equation*} x+2y=14 \end{equation}

    \begin{equation*} 15x-8y=20 \end{equation}

Then we proceed as before:

    \begin{equation*} (x+2y=14)\times 4 \implies 4x + 8y =56 \end{equation}

    \begin{equation*} (15x-8y=20)\times 1 \implies 15x-8y=20. \end{equation}

The last step can be skipped because multiplication by 1 doesn’t alter an equation. Now add to eliminate y:

    \[19x=76\implies x=4.\]

Substitute x=4 into equation (13) to find y:

    \[\frac{4}{2}+y=7, \quad\implies 2+y=7,\quad\implies y=5.\]

So x=4 and y=5. We leave you to verify this solution.

Example 8 (a word problem involving integers)

The sum of two numbers is 12, while their difference is 2. Find the two numbers.

  • Begin by assigning variables to represent the numbers; let’s choose x and y.
  • Understand the highlighted terms (sum and difference).
  • Formulate appropriate linear equations based on the given information.
  • Solve the resulting equations.

Following the above procedure, we’re led to the linear system

(15)   \begin{equation*} x+y=12 \end{equation*}

(16)   \begin{equation*} x-y=2 \end{equation*}

Add: \quad 2x=14,\quad \implies x=7.
Substitute to get y=5. Check that 7+5=12 and 7-5=2.

Don’t be afraid of word problems.

Example 9 (another integer related word problem)

The sum of the digits of a two-digit number is 10. If the digits are reversed, the original number decreases by 36. Find the number.

Let the digits of the two-digit number be x and y; for simplicity, let x be the tens digit and let y the ones digit. Then the actual number is 10x+y; but when the digits are reversed, the number becomes 10y+x. This, together with the information given in the question, leads to the linear system:

(17)   \begin{equation*} x+y=10 \end{equation*}

(18)   \begin{equation*} (10x+y)-(10y+x)=36 \end{equation*}

Equation (18) can be simplified as 9x-9y=36, or x-y=4. Our refactored equations are:



Add: 2x=14,\quad x=7. Substitute in equation (17) to find y:

    \[7+y=10,\quad y=3.\]

The original number is 73. Check that 7+3=10 and that 73-37=36.

Example 10 (still on numbers)

Two numbers are in the ratio of 1:2. If the numbers add up to 9, find the numbers.

Sometimes, a word problem is so easy that the solution can be obtained by inspection. Just like in this case. However, we want to use linear system to solve it because that’s our topic for now. So, let the numbers be x and y (assume that x is the smaller number). We have:
x:y=1:2 (or 2x=y, or 2x-y=0) and x+y=9.

(19)   \begin{equation*} 2x-y=0 \end{equation*}

(20)   \begin{equation*} x+y=9 \end{equation*}


So, x=3. Substitute in equation (19) to get 2(3)-y=0, whence y=6.
The numbers are 3 and 6. Check that 3:6=1:2, and 3+6=9, as required.

There are other class of word problems that generally lead to linear systems. For example, problems relating to speed/distance/time, problems relating to mixtures, and problems relating to money. And so on. You’ll find these in your textbook.

Graphical solution of linear systems

In order to solve a linear system using graphs, you must first of all be comfortable with graphing straight lines.

Review of linear graphs

Straight lines are very easy to graph. You just need to know the slope of the line and its y-intercept. Alternatively, you can also graph a straight line if you know both the x and y intercepts (actually, any two points will do, but using the intercepts makes it more convenient).

ExaMple 1

Sketch the graph of the line y=x-2.

Let’s use the intercept method to sketch this graph. To find the y-intercept, set x=0 and obtain y=0-2=-2; to find the x-intercept, set y=0 and solve for x:

    \[0=x-2,\quad x=2\]

Next, we join the y-intercept and the x-intercept with a straight line to obtain the graph of y=x-2 (red line below):

Rendered by

example 2

Sketch the graph of 2x+3y=6.

Again, let’s use the intercept method to sketch this graph.

  • x-intercept is obtained by putting y=0 and solving for x:

        \[2x+3(0)=6,\quad\implies 2x=6,\quad\implies x=3;\]

  • y-intercept is obtained by putting x=0 and solving for y:

        \[2(0)+3y=6,\quad\implies 3y=6,\quad\implies y=2.\]

We now join the x and y intercepts to obtain the graph below (blue line):

Rendered by

Example 3

Sketch the graph of 3x-2y-2=0.

Let’s use the slope and the y-intercept this time around. To do so, we first isolate the variable y as follows:

    \[3x-2 =2y,\quad\implies \frac{3x}{2}-\frac{2}{2}=\frac{2y}{2},\quad\implies \frac{3x}{2}-1=y.\]

The equation y=\frac{3x}{2}-1 is now in the form y=mx+b, so its slope is \frac{3}{2} and its y-intercept is -1.

  • Mark the point where y=-1 on the y-axis;
  • From that point (where y=-1), move three units up (which means that you’ll now be at y=2, since -1+3=2);
  • Now move two units to the right (because the slope is 3/2, we have to rise 3 units and run two units to the right. Remember that rise over run thing?)
  • If you did this correctly, the starting coordinate was (0,-1) and you should end up at the point (2,2). Connect these two points together to obtain the graph shown below (green line):

Rendered by

Linear systems via graphs

With the hope that the above examples on linear graphs suffice, we now move on to solving linear system problems using graphs.

example 4

Use graphs to solve the linear system

(1)   \begin{equation*} x+y=3 \end{equation*}

(2)   \begin{equation*} x-y=1 \end{equation*}

We first graph the two lines and notice where they intersect:

Rendered by

\boxed{\textrm{As shown in the graph above, the lines intersect at the point (2,1)}}

Example 5

Use graphical method to solve the linear system

(3)   \begin{equation*} 2x+5y=1 \end{equation*}

(4)   \begin{equation*} 6x-4y=3 \end{equation*}

Again, we graph the two lines and notice where they intersect:

Rendered by

\boxed{\textrm{The lines intersect at (0.5,0), which is the solution}.}
The example above reveals one major limitation of the graphical method, which has to do with determining the point of intersection when such a point does not have integer coordinates. Our example was chosen so that we can easily read the coordinates of the point of intersection. How about when such coordinates are unfamiliar fractions? Like (\frac{3}{7},\frac{5}{13})? Because of this, we do not recommend using the graphical method when the intersection point cannot be easily read from the graph, unless a question specifies so.

Example 6

Determine the point of intersection of the lines

(5)   \begin{equation*} x-y=2 \end{equation*}

(6)   \begin{equation*} 2x-2y=6 \end{equation*}

by sketching their graphs.

Rendered by

\boxed{\textrm{We see that the lines don't intersect; the system is inconsistent.}}
It appears that an inconsistent system is better understood when visualized by graphs. This is one power of the graphical method.

Example 7

Solve the linear system

(7)   \begin{equation*} x+2y=2 \end{equation*}

(8)   \begin{equation*} 2x+4y-4=0 \end{equation*}

As always, we first graph the two lines:

Rendered by

Oops \cdots. Only one line is visible — despite the fact that our program graphed two lines (beginning with the one with the red legend and then the one with the green legend). It happened in this case that equations (7) and (8) represent the same line; as such, they are said to be coincident. The number of solutions in this case is infinite.

Example 8

A barnyard has pigs and chickens. Altogether there are 28 legs and 18 eyes. Determine the number of each animal.

Let’s follow the strategy for solving word problems involving linear systems to address this question. Our first step is to introduce variables. So, suppose that the number of pigs is x and that the number of chickens is y.

  • A pig has four legs, while a chicken has two legs.
  • A pig has two eyes, just as a chicken has two eyes.
  • Since there are x pigs and each has four legs, the number of legs from the pigs is 4x.
  • Since there are y chickens and each has two legs, the number of legs contributed by the chickens is 2y.
  • Since the total number of legs is 28, it means that 4x+2y=28.
  • Since the total number of eyes is 18, it means that 2x+2y=18.

The above analysis is probably unnecessary. The whole point is that the question (and the variables we chose) lead to the linear system:

(9)   \begin{equation*} 4x+2y=28\quad\quad\implies 2x+y=14 \end{equation*}

(10)   \begin{equation*} 2x+2y=18\quad\quad\implies x+y=9 \end{equation*}

Because of the relatively big numbers involved, graphing is not recommended for this question. Nevertheless, we’ll go ahead and use graphs being that we’re discussing graphical method.

Rendered by

From the graph, the point of intersection is (5,4), which corresponds to 5 pigs and 4 chickens.

Example 9

A store sells four ballpoint pens and five pencils for 6.50, while five ballpoint pens and four pencils are sold for 7.00. Find how much a ballpoint pen costs and how much a pencil costs.

We’ve ignored the currency unit. Let the cost of a ballpoint pen be x. Let the cost of a pencil be y. We have the linear system:

(11)   \begin{equation*} 4x+5y=6.50 \end{equation*}

(12)   \begin{equation*} 5x+4y=7.00 \end{equation*}

And then the graphs:

Rendered by

From the graph, the point of intersection is (x,y)=(1,0.5). So, a ballpoint pen costs 1.00, while a pencil costs 0.50.

Example 10

Find two numbers whose sum is 6 and whose difference is 2.

Let the two numbers be x and y. Without loss of generality, suppose that y is the smaller number. Then:

(13)   \begin{equation*} x+y=6 \end{equation*}

(14)   \begin{equation*} x-y=2 \end{equation*}

The graphs are shown below:

Rendered by

From the graph, we see that x=4 and y=2. So the numbers are 4 and 2.