Problem 373 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ x y+3 y^{2}-x+4 y-7=0,2 x y+y^{2}-2 x-2 y+1=0 $$

Step-by-Step Solution

Answer

The solution to these system of non-linear equations would be set (pairs) of x and y values found.

Step 1: Rearrange first equation

First, we rearrange the first equation to find y in terms of x as follows: \(y(x) = (x+7) / (x+3)\)

Step 2: Substitute y(x) in second equation

Substitute \(y(x)\) from step 1 into the second equation. The second equation will become a function of x only.

Step 3: Simplify equation and Solve for x

After substituting y(x) in second equation solve the equation for x.

Step 4: Substitute x values to find corresponding y values

Once you have the value of x, you can substitute it back into the equation found for y(x) in Step 1 to find the corresponding y value. This gives the solution (x,y) to the system of equations.

Key Concepts

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. Its main role is to simplify the problem by reducing the number of variables in one equation. When faced with a system of two equations, as in our original problem, the goal is to express one variable in terms of the other.

• Step 1 involves isolating one of the variables in one equation to express it as a function of the other variable. In this case, we found an expression for \(y\) in terms of \(x\), \(y = \frac{x+7}{x+3}\).
• Step 2 involves substituting this expression into the other equation. This cleverly eliminates the variable \(y\), converting the system into a single-variable equation.

After substitution, we manage to focus our efforts on solving one equation, which is much more straightforward than dealing with two simultaneously. This is the essence of the substitution method: breaking down complex systems into simpler problems by substitution.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations so they become easier to solve. This is crucial in both setting up the substitution and in simplifying the equations thereafter. In the context of the original exercise, proper algebraic manipulation made it feasible to express \(y\) as \(y = \frac{x+7}{x+3}\).

• Rearrange terms so that terms involving the same variable are grouped together.
• Simplify expressions wherever possible, such as performing divisions, combining like terms, or factoring common elements.

These techniques often involve using addition, subtraction, multiplication, and division systematically and logically. Mastery of algebraic manipulation is essential for further steps in the solution, like transforming the second equation into one with a single variable. It turns a potentially intimidating problem into a series of manageable operations.

Equation Solving

Solving equations is the final step in many mathematical problems, and it involves finding values for variables that satisfy all equations in the system.

In our exercise, once we performed the substitution and algebraic manipulation, the focus spent on solving an equation solely in terms of \(x\). When you arrive at this point, look for values of \(x\) that satisfy the transformed equation. This may involve factoring, using the quadratic formula, or simply finding values that balance the equation.

• Example strategies include: factoring the equation to find \(x\), setting derivatives for optimization problems, and using numerical methods.
• The solution for \(x\) is then substituted back into the expression for \(y\) to find the corresponding \(y\) values.

It's key to remember that the solution gives a set of \((x,y)\) pairs, effectively solving the entire system. This step emphasizes careful handling and cross-verification to ensure solutions are correct and correspond well with the original equations given in the problem.