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.

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