Linear Equations in Two Variables

Linear Equations in Two Variables
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In this mini-lesson, we will explore solving a system of linear equations in two variables using different methods, solved examples, linear equations worksheets, and interactive questions.

The solution of a linear equation is the value(s) of the variable(s) which satisfies it.

For example, (2, 5) is a solution of \(2x-5y+21=0\) because:

\[2(2)-5(5)+21=0 \Rightarrow 0=0\]

But how to solve linear equations in two variables?

To solve the system of two linear equations in two variables, we have different methods.

You can explore the methods of solving linear equations, by entering any two linear equations in two variables and selecting the method.

Lesson Plan

What Are the Forms of Linear Equation?

A linear equation in two variables can be in different forms. Some of them are:

forms of linear equation flowchart

If you would like to know in detail about each form, click here.

The system of equations means the collection of equations.

We will learn how to solve linear equations in two variables using different methods.

Also read:

What are Linear Equations in One Variable?-Olympiad


Solving Linear Equations in two Variables (Graphically)

To solve a system of two equations in  variables graphically:

Just graph each equation.

To graph an equation manually,  first convert it to the form \(y=mx+b\) by solving the equation for \(y\) .

If you want to know how to graph a linear equation, see the following illustration.


If you are allowed to use a graphing calculator, then better using it to graph the linear equations.

Identify the point where both lines meet.

The point of intersection is the solution of the given system.

Example 

Let us find the solution of the following system of equations graphically.

\[ \begin{align} -x+2y-3&=0\\[0.2cm] 3x+4y-11&=0 \end{align} \]

We will graph them and see whether they intersect at a point.

Solving linear equations in 2 variables (Graphical): Given two equations are graphed which meet at (1, 2)

Both lines meet at (1, 2).

Thus, the solution of the given system of linear equations is:

\[x=1\\[0.1cm]y=2\]

But both lines always may not intersect. Sometimes they may be parallel. In that case, the given system has no solution.

In some other cases, both lines coincide with each other. In that case, each point on that line is a solution of the given system and hence the given system has an infinite number of solutions.

If the system has a solution, then it is said to be consistent; otherwise, it is said to be inconsistent. 

Solving Linear Equations in 2 Variables (Graphical): consistent, inconsistent, dependent and independent equations 

But how to determine each of these types algebraically? Let's see.

Consider a system of two linear equations:\[ \begin{align} a_1x+b_1y+c_1&=0\\[0.2cm] a_2x+b_2y+c_2&=0 \end{align} \]

Unique Solution
(Consistent and Independent)

\[\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\]

No Solution
(Inconsistent and Independent)

\[\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\]

Infinite Number of Solutions
(Consistent and Dependent)

\[\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\]

Method of Substitution

To solve a system of two equations in 2 variables using the substitution method

  • Solve one of the equations for one variable.
  • Substitute this in the other equation to get an equation in terms of a single variable.
  • Solve it for the variable.
  • Substitute it in any of the equations to get the value of another variable.

Example 

We will solve the following system of equations (that are labeled (1) and (2))  using the substitution method.

\[\begin{array}{ll}
x+2 y-7=0 & \cdots (1)\\[0.2cm]
2 x-5 y+13=0 & \cdots (2)
\end{array}\]

Let us solve the equation (1) for \(y\):

\[\begin{array}{l}
x+2 y-7=0 \\[0.2cm]
\Rightarrow \quad 2 y=7-x \\[0.2cm]
\Rightarrow \quad y=\dfrac{7-x}{2}\,\,\,\,\, \cdots (3)
\end{array}\]

Substitute this in the equation (2):

\[\begin{array}{l}
2 x-5 y+13=0 \\[0.2cm]
\Rightarrow \quad 2 x-5\left(\dfrac{7-x}{2}\right)+13=0 \\[0.2cm]
\Rightarrow \quad 2 x-\dfrac{35}{2}=\dfrac{5 x}{2}+13=0 \\[0.2cm]
\Rightarrow \quad 2 x+\dfrac{5 x}{2}=\dfrac{35}{2}-13 \\[0.2cm]
\Rightarrow \quad \dfrac{9 x}{2}=\dfrac{9}{2} \\[0.2cm]
\Rightarrow \quad x=1
\end{array}\]

Substitute this in the equation (3):

\[y=\dfrac{7-1}{2} = 3\]

Therefore, the solution of the given system is:

\[x=1\\[0.1cm]y=3\]


Cross Multiplication Method

Consider a system of linear equations:\[ \begin{align} a_1x+b_1y+c_1&=0\\[0.2cm] a_2x+b_2y+c_2&=0 \end{align} \]

To solve this using the cross multiplication method, we first write the coefficients of each of \(x\) and \(y\); and constants as follows:

system of linear equations by cross multiplication method

Here, the arrows indicate that those coefficients have to be multiplied.

Now we write the following equation by cross multiplying and subtracting the products.

\[\dfrac{x}{b_{1} c_{2}-b_{2} c_{1}}=\dfrac{y}{c_{1} a_{2}-c_{2} a_{1}}=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}}\]

From this equation, we get two equations:

\[\begin{align}
\dfrac{x}{b_{1} c_{2}-b_{2} c_{1}}&=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}} \\[0.2cm] \dfrac{y}{c_{1} a_{2}-c_{2} a_{1}}&=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}}
\end{align}\]

Solving each of these for \(x\) and \(y\), the solution of the given system is:

\(\begin{align}
x&=\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\[0.2cm] y&=\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}
\end{align}\)

Method of Elimination

To solve a system of two equations in 2 variables using the the elimination method

  • See if adding the equations would result in the cancellation of a variable.
  • If not, multiply one or both equations by some numbers such that their addition would result in the cancellation of a variable.
  • Solve the resulting single variable equation.
  • Substitute it in any of the equations to get the value of another variable.

Example 

We will solve the following system of equations using the elimination method.

\[\begin{array}{l}
2 x+3 y-11=0 \\
3 x+2 y-9=0
\end{array}\]

By adding these two equations would not result in the cancellation of any variable.

Let us aim at the cancellation of \(x\).

The coefficients of \(x\) in both equations are 2 and 3. Their LCM is 6

We will make the coefficients of \(x\) in both equations to be 6 and -6 so that the \(x\) terms get canceled when we add the equations.

\[\begin{array}{c}
\left\{\begin{array}{c}
3 \times(2 x+3 y-11=0) \\
-2 \times(3 x+2 y-9=0)
\end{array}\right. \\
\Rightarrow \quad 6 x+9 y-33=0 \\
-6 x-4 y+18=0
\end{array}\]

Now we will add these two equations:

\[\begin{array}{c}
\,6 x+9 y-33=0 \\
-6 x-4 y+18=0 \\
\hline 0+5 y-15=0\\ 
\Rightarrow \quad 5 y-15=0 \\ 
\Rightarrow \quad 5 y=15 \\ 
\Rightarrow \quad y=3 
\end{array}\]

Substitute this in one of the given two equations and solve the resultant variable for \(x\).

\[\begin{array}{l}
2 x+3 y-11=0 \\
\Rightarrow \quad 2 x+3(3)-11=0 \\
\Rightarrow \quad 2 x+9-11=0 \\
\Rightarrow \quad 2 x=2 \\
\Rightarrow \quad x=1
\end{array}\]

Therefore, the solution of the given system of equations is:

\[x=1\\[0.1cm]y=3\]


Determinants Method

The determinant of a 2 x 2 matrix is obtained by cross multiplying elements starting from the top left corner and subtracting the products.

Solving system of equations using determinants method

Consider a system of linear equations:\[ \begin{align} a_1x+b_1y&=c_1\\[0.2cm] a_2x+b_2y&=c_2 \end{align} \]

To solve them using the determinants method (which is also known as Crammer's Rule):

  • We first find the determinant formed by the coefficients of \(x\) and \(y\) and label it \(\Delta\).
    \[ \Delta = \left|\begin{array}{ll}a_1 & b_1 \\a_2 & b_2\end{array}\right| = a_1 b_2 - a_2b_1\]
  • We then find the determinant \(\Delta_x\) which is obtained by replacing the first column of \(\Delta\) with constants.
    \[ \Delta_x = \left|\begin{array}{ll}c_1 & b_1 \\c_2 & b_2\end{array}\right| = c_1 b_2 - c_2b_1\]
  • We then find the determinant \(\Delta_y\) which is obtained by replacing the second column of \(\Delta\) with constants.
    \[ \Delta_y = \left|\begin{array}{ll}a_1 & c_1 \\a_2 & c_2\end{array}\right| = a_1 c_2 - a_2c_1\]

Then the solution of the given system of linear equations is obtained by the formulas:

\( \begin{align} x &= \dfrac{\Delta_x}{\Delta}\\[0.2cm] y&= \dfrac{\Delta_y}{\Delta} \end{align}\)
 
tips and tricks
Tips and Tricks

While solving the equations using either the substitution method or the elimination method:

  • If we get an equation that is true (i.e., something like 0 = 0, -1 = -1, etc), then it means that the system has an infinite number of solutions.
  • If we get an equation that is false (i.e., something like 0 = 2, 3 = -1, etc), then it means that the system has no solution.

Solved Examples

Example 1

 

 

The sum of the digits of a two-digit number is 8. When the digits are reversed, the number is increased by 18. Find the number.

Solution

Let us assume that \(x\) and \(y\) are the tens digit and the ones digit of the required number.

Then the number is \(10x+y\).

The number when the digits are reversed is, \(10y+x\).

The problem says, "The sum of the digits of a two-digit number is 8"

So we get: \[x+y=8 \Rightarrow y=8-x\,\,\, \rightarrow (1)\]

Also, when the digits are reversed, the number is increased by 18. So the equation is:

\[ \begin{align}10y+x&=10x+y+18\\[0.2cm] 10(8-x)+x&=10x+(8-x)+18\,\,\, \text{(From (1))}\\[0.2cm] 80-10x+x&=10x+8-x+18\\[0.2cm] 80-9x&=9x+26\\[0.2cm] 18x &= 54\\[0.2cm] x&=3\end{align}\]

Substitute this in (1): \[y=8-3=5\]

So the required number is: \[10x+y=10(3)+5 =35\]

35
Example 2

 

 

Jake's piggy bank has11 coins (only quarters or dimes) that have a value of S1.85. How many dimes and quarters does the piggy bank has?

Solution

Let us assume that

  • the number of dimes = \(x\)
  • the number of quarters = \(y\)

Since there are 11 coins in total,  \[x+y=11 \Rightarrow y=11-x\,\,\, \rightarrow (1)\]

The total value of the money in the piggy bank is $1.85 (185 cents). So we get the equation:

\[ \begin{align}10x+25y&=185\\[0.2cm] 10x + 25(11-x)&=185\,\,\, \text{(From (1))}\\[0.2cm] 10x+275-25x &=185\\[0.2cm] -15x +275&=185\\[0.2cm] -15x&=-90\\[0.2cm] x&=6\end{align}\]

Substitute this in (1): \[ y=11-6=5\]

Therefore:

The number of dimes = 6
The number of quarters = 5
Example 3

 

 

In a river, a boat can travel 30 miles upstream in 2 hours. The same boat can travel 51 miles downstream in 3 hours. Then

  • What is the speed of the boat in still water? 
  • What is the speed of the current?

Solution

Let us assume that:

  • the speed of boat in still water = \(x\) miles per hour
  • the speed of current = \(y\) miles per hour.

During upstream, the current pulls back the boat's speed and the speed of the boat during upstream = \(x-y\).

During downstream, the current's speed adds to the boat's speed and the speed of the boat during downstream = \(x+y\).

Thus,

 

Distance
\(d\)

Time
\(t\)
Speed Speed
= d / t
Upstream 30 2 \(x-y\) 30/2=15
Downstream 51 3 \(x+y\) 51/3=17

Using the last two columns of the table:

\[ \begin{align} x-y&=15 \,\,\, \rightarrow (1)\\[0.2cm] x+y &=17\,\,\, \rightarrow (2) \end{align}\]

Adding (1) and (2):

\[\begin{align} 2x &= 32\\[0.2cm] x&=16 \end{align}\]

Substitute this in (2):

\[\begin{align} 16+y&= 17\\[0.2cm] y&=1 \end{align}\]

Therefore,

The speed of boat = 16 miles per hour
The speed of current = 1 mile per hour

You can practice the linear equations worksheets that are given at the bottom of this page to become perfect in this topic.

 
Challenge your math skills
Challenging Questions
  1. An airplane flying with the wind can cover 500 miles in 4 hours.  The return trip against the wind takes 5.5 hours. Find the speed of the plane and the speed of the wind.
  2. Solve the following equations:
    \[ \begin{align}
     \dfrac{1}{x}+ \dfrac{3}{y}&= 7\\[0.2cm]
     \dfrac{2}{x}- \dfrac{5}{y}&= 8
    \end{align}\]
    Hint: Assume \( \dfrac{1}{x} = a\) and \( \dfrac{1}{y}=b\).

Interactive Questions

Here are a few activities for you to practice.

Select/type your answer and click the "Check Answer" button to see the result.

 
 
 
 
 
 

Let's Summarize

The mini-lesson targeted the fascinating concept of Linear Equations in Two Variables. The math journey around Linear Equations in Two Variables starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

About Cuemath

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.


Frequently Asked Questions (FAQs)

1. What are linear equations?

A linear equation is an equation in which the variable(s) is(are) with the exponent 1.

Example: 

\[ \begin{align} 2x &=45\\ x+y&=35\\ a-b &= 45 \end{align}\]

2. How to graphically represent linear equations?

We convert the linear equation to the slope-intercept form to graph it. If you want to learn about this in detail, view the following simulation.

3. How does one solve the system of linear equations?

We have different methods to solve the system of linear equations:

You can learn about each method in detail on this page.

Download Pair of Linear Equations in 2 Variables Worksheets
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 1
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 1
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 2
  
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