LINEAR EQUATIONS IN TWO VARIABLES-Notes
Maths - Notes
Linear Equation
A linear equation is a mathematical equation in which the highest power (degree) of the variable is always 1, making it a first-degree equation. Linear equations can have one or more variables, and their graph always forms a straight line.
Definition and Standard Forms
- In one variable, a linear equation is written as \(ax+b=0\), where \(a,\;b\) are constants, and \(x\) is the variable.
- In two variables, the standard form is \(ax+by+c=0\), where \(x\) and \(y\) are variables and \(a,\,b,\,c\) are constants.
- More generally, for \(n\) variables, it can be expressed as \(a_1x_1 + a_2x_2 + \cdots + a_nx_n + b=0\), with at least one \(a_i \neq 0\).
Characteristics
- No variable in a linear equation has an exponent greater than 1, and variables are not multiplied together.
- The graph of a linear equation is always a straight line in two dimensions or a hyperplane in higher dimensions.
- Solutions to a linear equation are the values of the variables that make the equation true.
Forms of Representation
-
Standard Form:
\(ax+b=0\) (one variable), \(ax+by+c=0\) (two variables), \(ax+by+cz+d=0\) (three variables). - Slope-Intercept Form:
\(y=mx+c\), where \(m\) is the slope, and \(c\) is the y-intercept. - Point-Slope Form:
Useful when the slope and a point on the line are known.
Examples
- \(2x-3=0\) is a linear equation in one
variable.
Graphical Representation of \(p(x)=2x-3=0\) - \(3x+4y=12\) is a linear equation in two
variables
Graphical Representation of \(3x+4y=12\) - \(x+y+z=7\) is a linear equation in three variables.
Real-Life Application
Linear equations are used widely in mathematics, science, and engineering because many physical systems can
be approximated by linear relationships.
A linear equation serves as a foundational concept in algebra, ensuring that the relationship between
variables is direct and proportional.
Methods of Solution of a linear equation in one variable
To solve a linear equation in one variable, follow a systematic process that uses basic algebraic operations to isolate the variable and find its value.
Step-by-Step Method
- Step 1: Simplify both sides:
If there are fractions, clear them by multiplying all terms by the least common multiple (LCM) of denominators.
Combine like terms and expand parentheses if needed. - Step 2: Bring variable terms to one side:
Use addition or subtraction to collect all terms containing the variable on one side, and constants on the other side. - Step 3: Isolate the variable:
Use multiplication or division (the properties of equality) to make the coefficient of the variable equal to 1. - Step 4: Check your solution:
Substitute your solution back into the original equation to verify it makes both sides equal.
Step-by-step solving for \(5x - 9 = -3x + 19\)
Solve \(5x - 9 = -3x + 19\)
Move all variable terms to one side
\[\begin{aligned}5x+3x−9&=19\\8x−9&=19\end{aligned}\]
Move constants to the other side:
\[\begin{aligned}8x&=19+9\\8x&=28\end{aligned}\]
Divide by the coefficient:
\[\begin{aligned}x&=\frac{28}{8}\\\\
x&=3.5\end{aligned}\]
Verify the solution by substituting x into the original equation
To verify the solution:
x=3.5 for the equation \(5x−9=−3x+19\), substitute x=3.5 into the original equation:
Original equation:
\[5x−9=−3x+19\]
Substitute \(x=3.5\):
\[5×3.5−9=−3×3.5+19\]
Calculate each side:
\[\begin{aligned}17.5−9&=−10.5+19\\8.5&=8.5\end{aligned}\]
Since both sides are equal, the solution \(x=3.5\) is correct.
Prove that substitution works for all linear equations
To prove that substitution works for all linear equations, consider the fundamental properties of equality
and the structure of linear equations:
Linear Equation Form
A linear equation in one variable can be written as:
\[ax+b=0\] where \(a\) and \(b\) are constants and \(a\neq0\)
Solving Using Substitution
The solution is found by isolating \(x\)
\[x=-\frac{b}{a}\]
Proof that Substitution Always Works
-
Definition of Solution:
Substituting \[x=−\frac{b}{a}\] back into the equation must satisfy the equality: \[a\left(-\frac{b}{a}\right)+b=0\] - Evaluate Substitute:
\[-b+b=0\]
This simplifies to zero, satisfying the equation.
- Property of Equality:
The substitution replaces\(x\) with the value that makes the equation true by directly negating the constant term with respect to the coefficient. - Generality:
Because linear equations always have this form and the substitution exactly balances the equation, it follows that substitution will always verify or confirm the solution for any linear equation in one variable.
This proof holds for all linear equations due to their simple first-degree structure and the equivalence
property of equality.
Methods of solving simulteneous Equations
Simultaneous equations are two or more equations involving the same variables that must be satisfied at the same time. The goal is to find values of the variables that make all equations true simultaneously. There are several reliable methods to solve them, each suitable for different situations.
- Substitution Method:
- Elimination Method:
- Graphical Method:
- Cross-Multiplication Method:
Method of Elimination by Substitution
In this method, one equation is rearranged to express one variable in terms of the other, and that
expression is substituted into the other equation.
Steps:
- Rearrange one equation to make one variable the subject.
- Substitute this expression into the other equation.
- Solve for one variable.
- Substitute back to find the second variable.
Example:
\[3x−4y=0\tag{1}\] and \[9x−8y=12\tag{2}\]From the first equation:
\[x=\frac{4y}{3}\] Substitute into the second: \[\require{cancel} \begin{aligned} 9x - 8y &= 12 \\ 9\left(\frac{4y}{3}\right) - 8y &= 12 \\ \frac{\cancelto{3}{9} \times 4y}{\cancelto{1}{3}} - 8y &= 12 \\ 12y - 8y &= 12 \\ 4y &= 12 \\\\ y &= \frac{\cancelto{3}{12}}{\cancelto{1}{4}} \\\\ y &= 3 \end{aligned} \]
Now substitute \(y=3\) into \(x=\frac{4y}{3}\)
\[\begin{aligned}x&=\frac{4\times3}{3}\\\\&=4\end{aligned}\]
Solution: \(x=4\), \(y=3\)
Method of Elimination by Equating Coefficients
Steps:
- Multiply one or both equations so that coefficients of one variable are equal.
- Add or subtract the equations to eliminate that variable.
- Solve for the remaining variable.
- Substitute back to find the other variable.
Example:
\[4a+5b=12\tag{1}\] and \[3a−5b=9\tag{2}\] add both equations:\[\begin{aligned} 4a+5b+3a−5b&=12+9\\ 4a+3a+5b-5b&=21\\ 7a&=21\\ &=3 \end{aligned}\] Substitute \(a=3\) into the first equation: \[\begin{aligned} 4a+5b&=12\\ 4\times3 +5b&=12\\ 12+5b&=12\\ 5b&=12-12\\ 5b&=0\\ \implies b&=0 \end{aligned}\]
Solution: \(a=3\), \(b=0\).
Graphical Method
Example
\[x+y=5\tag{1}\] and \[2x−y=4\tag{2}\] Writing both equation in form \(y=mx+c\) \[\begin{aligned}x+y&=5\\y&=5-x\end{aligned}\tag{3}\] \[\begin{aligned}2x−y&=4\\y&=2x-4\end{aligned}\tag{4}\] Plotting Graphs of equation 3 and 4
By graphing both lines, they intersect at\((3,2)\)
Solution: \(x=3\) and \(y=2\)
Method of Cross Multiplication
\[a_1x+b_1y+c_1\tag{1}\] and \[a_2x+b_2y+c_2\tag{2}\] The Formula is: \[\scriptsize\boxed{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}}\]
Methods of Solving Simultaneous Equations
| Method | Key Idea | When to Use | Example Result |
|---|---|---|---|
| Substitution | Replace one variable using another | Easy when one coefficient is 1 | x = 4, y = 3 |
| Elimination | Add/subtract equations to cancel one variable | When coefficients can be adjusted | a = 3, b = 0 |
| Graphical | Plot both lines and find intersection | Visual and conceptual | (3, 2) |
| Cross-Multiplication | Use determinant-like ratios | Quick for two-variable linear systems | Calculated algebraically |
Example-1
Solve the following system of equations using the method of elimination by substitution.
\(x+y=7\) and \(3x-2y=11\)
Solution:
$$
\begin{aligned}
&\text{Given equations:} \\
x + y &= 7 \\
3x - 2y &= 11 \\[6pt]
\end{aligned}
$$
From the first equation, express \(x\) in terms of \(y\):
$$x = 7 - y $$
Substitute this value of \(x\) into the second equation:
$$
\begin{aligned}
3(7 - y) - 2y &= 11 \\[6pt]
21 - 3y - 2y &= 11 \\[6pt]\end{aligned}$$
Combine like terms:
$$\begin{aligned}-5y &= 11 - 21 \\[6pt]
-5y &= -10 \\[6pt]
y &= 2 \\[6pt]\end{aligned}$$
Substitute \(y = 2\) into \(x = 7 - y\):
$$\begin{aligned}x &= 7 - 2 \\[6pt]
x &= 5
\end{aligned}
$$
Example-2
Solve using elimination bt substitution:
\[\scriptsize\frac{x+7}{5} - \frac{2x-y}{4}=3y-5\] and \[\scriptsize\frac{4x-3}{6}+\frac{5y-7}{2}=18-5x\]
Solution:
$$\small\begin{aligned}\frac{x+7}{5}-\frac{2x-y}{4}&=3y-5\text{ and }\\\\ \frac{4x-3}{6}+\frac{5y-7}{2}&=18-5x\\\\ \frac{x+7}{5}-\frac{\left( 2x-y\right) }{9}&=3y-5\\\\ \frac{4x+28-10x+5y}{20}&=3y-5\\\\ 4x-10x+5y+28&=60y-100\\ -6x+5y-60y&=-100-28\\ -6x-55y&=-128\\ -6x&=55y-128\\ x&=\frac{128-55y}{6}\\\\ \frac{4x-3}{6}+\frac{5y-7}{2}&=18-5x\\\\ \Rightarrow \frac{4x-3+15y-21}{6}&=18-5x\\\\ \Rightarrow 4x+15y-24&=108-30x\\ \Rightarrow 4x+30x+15y&=108+24\\ \Rightarrow 34x+15y&=132\\ \Rightarrow 34\times \left( \frac{128-55y}{6}\right) +15y&=132\\\\ \Rightarrow \frac{17}{3}\times \left( 128-55y\right) +15y&=132\\\\ \Rightarrow 17\left( 128-55y\right) +45y&=396\\ \Rightarrow 2176-935y+45y&=396\\ \Rightarrow 2176-890y-396&=0\\ \Rightarrow 890y&=2176-396\\ \Rightarrow 890y&=1780\\ \Rightarrow y&=\frac{1780}{890}\\\\ &=2\\ y&=2\\\\ x&=\frac{128-55y}{6}\\ x&=\frac{128-55\times 2}{6}\\\\ &=\frac{128-110}{6}\\\\ &=\frac{18}{6}\\\\ &=3\\ x=3,\;y=2\end{aligned}$$
Example-3
Solve, uing the method of elimination by equating coefficient: \[3x-4y=10\] and \[5x-3y=24\]
Solution:
$$\begin{align}3x-4y&=10\tag{1}\\
5x-3y&=24\tag{2}\end{align}$$
Multiply equation (1) by 5 and equation(2) by 3
$$\begin{aligned}3x-4y&=10\quad\cdots\times 5\\
5x-3y&=24\quad\cdots\times 3\\\\
15x-20y&=50\\
15x-\;\;9y&=72\\
\color{red}-\quad\quad+\quad&\quad\color{red}-\\\hline
-11y&=-22\\\\
y=\frac{22}{11}\\\\
y=2\end{aligned}$$Substituting value of \(y=2\) in eqn (1)
$$\begin{aligned}3x-4\times 2&=10\\
3x-8&=10\\
3x&=18\\
x&=\frac{18}{3}\\
&=6\\
\boxed{x=6,\;y=2}\end{aligned}$$
Example-4
Solve by Cross-Multiplication \[3x+y=13\]and \[x-3y+9=0\]
Solution:
$$\scriptsize\begin{aligned}3x+y&=13\\ x-3y+9&=0\end{aligned}$$ $$\scriptsize\frac{x}{\begin{vmatrix} b_{1} & c_{1} \\ b_{2} & c_{2} \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_{1} & a_{1} \\ c_{2} & a_{2} \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}}\\\\ \scriptsize\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{y}{c_{1}a_{2}-c_{2}a_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\\$$ $$\scriptsize\begin{aligned}a_{1}&=3\\a_{2}&=1\\ b_{1}&=1\\b_{2}&=-3\\ c_{1}&=-13\\c_{2}&=9\\\\ \frac{x}{9-39}&=\frac{y}{-13-27}=\frac{1}{-9-1}\\\\ \frac{x}{-30}&=\frac{y}{-40}=\frac{1}{-10}\\\\ \frac{x}{-30}&=\frac{1}{-10}\\\\ x&=\frac{30}{10}\\\\ x&=3\\\\ \frac{y}{-40}&=\frac{1}{-10}\\\\ y&=\frac{40}{10}\\\\ y&=4\\\\ \quad\boxed{x=3,\;y=4}\end{aligned}$$
Example-5
Solve:\[\frac{7}{x}+\frac{8}{y}=2\] and \[\frac{2}{x}+\frac{13}{y}=22\]
Solution
$$\begin{align}\frac{7}{x}+\frac{8}{y}&=2\tag{A}\\\\
\frac{2}{x}+\frac{13}{y}&=22\tag{B}\\\\
\frac{7}{x}+\frac{8}{y}&=2\\\\
\Rightarrow 7y+8x&=2xy\tag{1}\\\\
\frac{2}{x}+\frac{13}{y}&=22\\\\
\Rightarrow 2y+13x&=22xy\tag{2}\\\end{align}$$
Dividing Equation(1) by Equation(2)
$$\begin{align}\frac{7y+8x}{2y+13x}&=\frac{2xy}{22xy}\\\\
\Rightarrow \frac{7y+8x}{2y+13x}&=\frac{1}{11}\\\\
\Rightarrow \left( 7y+8x\right) \times 11&=2y+13x\\\\
\Rightarrow 77y+88x-2y-13x&=0\\\\
\Rightarrow 75y+75x&=0\\\\
\Rightarrow y&=-x\end{align}$$
Substituting the vale uf \(y\) in Equation(A)
$$\begin{align}\frac{7}{x}+\frac{8}{y}&=2\\\\
\frac{7}{x}+\frac{8}{-x}&=2\\\\
\frac{7-8}{x}&=2\\\\
\Rightarrow 2x&=-1\\\\
\Rightarrow x&=-1/2\\\\
y&=1/2\end{align}$$
