Equations Reducible to Linear Form

Equations reducible to a pair of linear equations are equations that, through a suitable substitution, can be transformed into a system of two linear equations in two variables. These linear equations can then be solved using various methods like substitution, elimination, or graphical methods. The original equation might appear non-linear (e.g., involving fractions or radicals), but the key is to identify a pattern and substitute appropriately to simplify it into a manageable system of linear equations. The goal is to rewrite a complex equation into a more familiar and solvable form.

Formulae

There isn’t a single formula for these types of equations. The strategy relies more on recognizing patterns and making appropriate substitutions. However, the general idea involves these steps:

• Identify the pattern: Look for repeating expressions or common factors within the equation.
• Substitute: Replace the repeating expression with a new variable (e.g., $u$, $v$, $p$, $q$).
• Simplify: Rewrite the equation in terms of the new variables. This should result in a pair of linear equations.
• Solve the linear system: Use your preferred method (substitution, elimination, etc.) to find the values of the new variables.
• Back-substitute: Replace the new variables with their original expressions and solve for the original variables (usually $x$ and $y$).

Examples

Here are two examples demonstrating how to solve equations reducible to a pair of linear equations:

Example-1

Solve the following equations:

$\frac{5}{x-1} + \frac{1}{y-2} = 2$
$\frac{6}{x-1} – \frac{3}{y-2} = 1$

Solution:

Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$. Then the equations become:
$5u + v = 2$ …(1)
$6u – 3v = 1$ …(2)

Multiply equation (1) by 3: $15u + 3v = 6$ …(3)

Add equations (2) and (3): $21u = 7$, so $u = \frac{1}{3}$.

Substitute $u = \frac{1}{3}$ into equation (1): $5(\frac{1}{3}) + v = 2$, so $v = 2 – \frac{5}{3} = \frac{1}{3}$.

Now, substitute back:
$u = \frac{1}{x-1} = \frac{1}{3} \Rightarrow x-1 = 3 \Rightarrow x = 4$
$v = \frac{1}{y-2} = \frac{1}{3} \Rightarrow y-2 = 3 \Rightarrow y = 5$

Therefore, the solution is $x = 4$ and $y = 5$.

Example-2

Solve the following equations:

$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$
$\frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1$

Solution:

Let $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$. The equations become:
$2u + 3v = 2$ …(1)
$4u – 9v = -1$ …(2)

Multiply equation (1) by 3: $6u + 9v = 6$ …(3)

Add equations (2) and (3): $10u = 5$, so $u = \frac{1}{2}$.

Substitute $u = \frac{1}{2}$ into equation (1): $2(\frac{1}{2}) + 3v = 2$, so $1 + 3v = 2$, and $v = \frac{1}{3}$.

Now, substitute back:
$u = \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$
$v = \frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9$

Therefore, the solution is $x = 4$ and $y = 9$.

Common mistakes by students

Students often make the following mistakes:

• Incorrect Substitution: Failing to identify the appropriate expression to substitute. Carefully examine the equation for repeating patterns or common factors.
• Algebraic Errors: Making mistakes when solving the resulting linear system. Double-check all calculations.
• Forgetting to Back-Substitute: Solving for the new variables (e.g., $u$ and $v$) but forgetting to find the values of the original variables (e.g., $x$ and $y$).
• Incorrectly Simplifying Fractions: When dealing with fractional expressions, make sure to simplify the expressions correctly.

Real Life Application

Equations reducible to a pair of linear equations can model various real-life situations. While the direct application might be less obvious than simple linear equations, the underlying principles are applicable to:

• Mixture Problems: Similar to the examples, problems involving mixtures of two substances can sometimes be set up with fractional equations.
• Rate Problems: Problems involving rates of work or rates of travel can sometimes be translated into equations that become linear after substitution.
• Optimization Problems (indirectly): While the final solution might be found using other methods, the initial setup of an optimization problem might involve equations that are initially non-linear but can be transformed.

Fun Fact

The method of substitution used in these types of problems has roots in early algebra techniques. It’s a fundamental concept that lays the groundwork for more advanced algebraic manipulations and problem-solving strategies. The ability to recognize and simplify complex expressions is a valuable skill across many areas of mathematics and beyond.

Recommended YouTube Videos for Deeper Understanding

Q.1 If $\frac{2}{x} + \frac{3}{y} = 13$ and $\frac{5}{x} – \frac{4}{y} = -2$, then find the values of $x$ and $y$.

Check Solution

Ans: A

Let $\frac{1}{x} = u$ and $\frac{1}{y} = v$. Then the equations become $2u + 3v = 13$ and $5u – 4v = -2$. Multiply the first equation by 4 and the second equation by 3. We get $8u + 12v = 52$ and $15u – 12v = -6$. Adding these gives $23u = 46$, so $u = 2$. Substituting $u=2$ in $2u + 3v = 13$ gives $4 + 3v = 13$, so $3v = 9$, and $v = 3$. Therefore, $x = \frac{1}{u} = \frac{1}{2}$ and $y = \frac{1}{v} = \frac{1}{3}$.

Q.2 Solve for $x$ and $y$: $\frac{5}{x-1} + \frac{1}{y-2} = 2$ and $\frac{6}{x-1} – \frac{3}{y-2} = 1$.

Check Solution

Ans: A

Let $\frac{1}{x-1} = u$ and $\frac{1}{y-2} = v$. The equations become $5u + v = 2$ and $6u – 3v = 1$. Multiply the first equation by 3: $15u + 3v = 6$. Add this to $6u – 3v = 1$ to get $21u = 7$, so $u = \frac{1}{3}$. Then $v = 2 – 5u = 2 – \frac{5}{3} = \frac{1}{3}$. Thus, $x-1=3$, so $x=4$, and $y-2=3$, so $y=5$.

Q.3 The equations $\frac{a}{x} – \frac{b}{y} = 0$ and $\frac{a b^2}{x} + \frac{a^2 b}{y} = a^2 + b^2$ have the solution $x = 1, y = -1$. What are the values of $a$ and $b$?

Check Solution

Ans: C

Substitute $x = 1$ and $y = -1$ into the given equations: $a – (-b) = 0$ and $ab^2 + a^2b = a^2 + b^2$. From the first equation, $a + b = 0$, so $b = -a$. Substituting into the second equation gives $a(-a)^2 + a^2(-a) = a^2 + (-a)^2$, which means $a^3 – a^3 = 2a^2$. Thus, $2a^2=0$, and so $a = 0$. But then $b = -a = 0$. Checking these values into original equations. If $a = 1$, $b = -1$, the equation becomes $\frac{1}{x} – \frac{-1}{y}=0$, so $\frac{1}{x} + \frac{1}{y} = 0$ and $\frac{-1}{x} + \frac{1}{y} = 2$. Using $x=1$, $y=-1$ satisfies the first equation: $\frac{1}{1} + \frac{1}{-1} = 0$. The second equation gives $\frac{1}{1} + \frac{-1}{-1}=2$, which implies $1+1 = 2$.

Q.4 If $\frac{x+y}{xy} = 2$ and $\frac{x-y}{xy} = 6$, find the values of $x$ and $y$.

Check Solution

Ans: A

Rewrite the equations as $\frac{x}{xy} + \frac{y}{xy} = 2$ and $\frac{x}{xy} – \frac{y}{xy} = 6$. This simplifies to $\frac{1}{y} + \frac{1}{x} = 2$ and $\frac{1}{y} – \frac{1}{x} = 6$. Adding these two equations gives $\frac{2}{y} = 8$, so $y = \frac{1}{4}$. Substituting $y = \frac{1}{4}$ in $\frac{1}{x} + \frac{1}{y} = 2$ gives $\frac{1}{x} + 4 = 2$, so $\frac{1}{x} = -2$, and $x = -\frac{1}{2}$.

Q.5 Solve for $x$ and $y$: $\frac{10}{x+y} + \frac{2}{x-y} = 4$ and $\frac{15}{x+y} – \frac{5}{x-y} = -2$.

Check Solution

Ans: A

Let $\frac{1}{x+y} = u$ and $\frac{1}{x-y} = v$. Then the equations become $10u + 2v = 4$ and $15u – 5v = -2$. Divide the first equation by 2: $5u + v = 2$. Multiply this by 5: $25u + 5v = 10$. Add this to $15u – 5v = -2$. Thus $40u = 8$, so $u = \frac{1}{5}$. Then $v = 2 – 5u = 2 – 1 = 1$. Therefore, $x+y=5$ and $x-y=1$. Adding these gives $2x=6$, so $x=3$. Substituting $x=3$ in $x+y=5$ gives $3+y=5$, so $y=2$.

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