If $\frac{x}{a} + \frac{y}{b} = a + b$ and $\frac{x}{a^2} + \frac{y}{b^2} = 2$ , then what is $\frac{x}{a^2} - \frac{y}{b^2}$ equal to?

Question

If $\frac{x}{a} + \frac{y}{b} = a + b$  and  $\frac{x}{a^2} + \frac{y}{b^2} = 2$ , then what is  $\frac{x}{a^2} - \frac{y}{b^2}$  equal to?

Prepp Team · Accepted Answer

Solving the System of Equations

We are given a system of two linear equations involving variables $x$ and $y$, along with parameters $a$ and $b$. The equations are:

$\frac{x}{a} + \frac{y}{b} = a + b$
$\frac{x}{a^2} + \frac{y}{b^2} = 2$

We need to find the value of the expression $\frac{x}{a^2} - \frac{y}{b^2}$.

To make the problem easier to handle, let's introduce new variables based on the expression we need to find. Let:

$P = \frac{x}{a^2}$
$Q = \frac{y}{b^2}$

Our goal is to find the value of $P - Q$.

Let's rewrite the given equations in terms of $P$ and $Q$.

Equation (2) is already in terms of $P$ and $Q$:

$P + Q = 2$ (Equation 3)

Now let's look at Equation (1): $\frac{x}{a} + \frac{y}{b} = a + b$.

We can express $\frac{x}{a}$ in terms of $P$:

$\frac{x}{a} = a \cdot \left(\frac{x}{a^2}\right) = aP$

Similarly, we can express $\frac{y}{b}$ in terms of $Q$:

$\frac{y}{b} = b \cdot \left(\frac{y}{b^2}\right) = bQ$

Substitute these into Equation (1):

$aP + bQ = a + b$ (Equation 4)

Now we have a system of two linear equations in the variables $P$ and $Q$:

$aP + bQ = a + b$
$P + Q = 2$

We can solve this system for $P$ and $Q$. Let's use the elimination method. Multiply the second equation ($P + Q = 2$) by $b$:

$b(P + Q) = b(2)$
$bP + bQ = 2b$ (Equation 5)

Now, subtract Equation (5) from Equation (4):

$(aP + bQ) - (bP + bQ) = (a + b) - 2b$
$aP + bQ - bP - bQ = a + b - 2b$
$aP - bP = a - b$
$P(a - b) = a - b$

Assuming $a - b \neq 0$, we can divide both sides by $(a - b)$:

$P = \frac{a - b}{a - b}$
$P = 1$

Now that we have the value of $P$, substitute it back into Equation (3) ($P + Q = 2$) to find $Q$:

$1 + Q = 2$
$Q = 2 - 1$
$Q = 1$

So, we found that $P = 1$ and $Q = 1$. Remember that $P = \frac{x}{a^2}$ and $Q = \frac{y}{b^2}$. Therefore:

$\frac{x}{a^2} = 1$
$\frac{y}{b^2} = 1$

We are asked to find the value of $\frac{x}{a^2} - \frac{y}{b^2}$, which is $P - Q$.

$P - Q = 1 - 1$
$P - Q = 0$

Thus, the value of $\frac{x}{a^2} - \frac{y}{b^2}$ is 0.

Step-by-Step Calculation

Identify the given system of equations.
Define substitution variables $P = \frac{x}{a^2}$ and $Q = \frac{y}{b^2}$.
Rewrite the original equations in terms of $P$ and $Q$:
- From $\frac{x}{a} + \frac{y}{b} = a + b$, substitute $\frac{x}{a} = aP$ and $\frac{y}{b} = bQ$ to get $aP + bQ = a + b$.
- From $\frac{x}{a^2} + \frac{y}{b^2} = 2$, substitute $P = \frac{x}{a^2}$ and $Q = \frac{y}{b^2}$ to get $P + Q = 2$.
Solve the system of equations for $P$ and $Q$:
- Use elimination or substitution. Multiply $P+Q=2$ by $b$ to get $bP+bQ=2b$.
- Subtract $bP+bQ=2b$ from $aP+bQ=a+b$ to get $P(a-b) = a-b$.
- Assuming $a \neq b$, solve for $P$: $P = 1$.
- Substitute $P=1$ into $P+Q=2$ to solve for $Q$: $1+Q=2 \implies Q=1$.
Calculate the required expression $P - Q$: $1 - 1 = 0$.

Equation	Transformation	In terms of $P$ and $Q$
$\frac{x}{a} + \frac{y}{b} = a + b$	$\frac{x}{a} = a \cdot \frac{x}{a^2}$, $\frac{y}{b} = b \cdot \frac{y}{b^2}$	$aP + bQ = a + b$
$\frac{x}{a^2} + \frac{y}{b^2} = 2$	Substitute $P = \frac{x}{a^2}$, $Q = \frac{y}{b^2}$	$P + Q = 2$

System in $P, Q$	Method	Result
$aP + bQ = a + b$ $P + Q = 2$	Solve for $P$ (assuming $a \neq b$)	$P = 1$
$P + Q = 2$	Substitute $P=1$, Solve for $Q$	$Q = 1$
Find $P - Q$	Calculate $1 - 1$	$0$

The value of $\frac{x}{a^2} - \frac{y}{b^2}$ is 0.

Revision Table: System of Equations

Reviewing key concepts related to solving systems of linear equations.

A system of linear equations is a set of two or more linear equations involving the same variables.
A solution to a system is a set of values for the variables that satisfy all equations simultaneously.
Methods for solving systems include substitution, elimination, and graphical methods.
Substitution involves solving one equation for a variable and substituting that expression into the other equation.
Elimination involves multiplying equations by constants and adding or subtracting them to eliminate a variable.
For a system of two linear equations in two variables, there can be a unique solution, no solution, or infinitely many solutions.
The case of a unique solution typically occurs when the lines represented by the equations are not parallel and distinct. In algebraic terms, this often happens when the determinant of the coefficient matrix is non-zero. For the system $aP + bQ = a + b$ and $P + Q = 2$, the coefficients of $P$ and $Q$ are $\begin{pmatrix} a & b \\ 1 & 1 \end{pmatrix}$. The determinant is $a(1) - b(1) = a - b$. A unique solution exists if $a - b \neq 0$, i.e., $a \neq b$.

Additional Information: Algebraic Manipulation

Algebraic manipulation is crucial for solving equations and simplifying expressions. Key techniques used here include:

Rearranging terms: Moving terms from one side of the equation to the other while changing their sign.
Factoring: Expressing a polynomial as a product of factors, like factoring out $P$ from $aP - bP$ to get $P(a - b)$.
Substitution: Replacing a variable or expression with another equivalent expression. We used this by defining $P$ and $Q$ and substituting them into the equations.
Solving for a variable: Isolating a variable on one side of the equation.

When working with equations involving fractions, multiplying by the least common multiple of the denominators can help clear the fractions, although in this case, direct substitution after defining $P$ and $Q$ was effective.

Prepp Team · Answer

-2

Prepp Team · Answer

-1

Prepp Team · Answer

1

Equation	Transformation	In terms of \(P\) and \(Q\)
\(\frac{x}{a} + \frac{y}{b} = a + b\)	\(\frac{x}{a} = a \cdot \frac{x}{a^2}\), \(\frac{y}{b} = b \cdot \frac{y}{b^2}\)	\(aP + bQ = a + b\)
\(\frac{x}{a^2} + \frac{y}{b^2} = 2\)	Substitute \(P = \frac{x}{a^2}\), \(Q = \frac{y}{b^2}\)	\(P + Q = 2\)

System in \(P, Q\)	Method	Result
\(aP + bQ = a + b\) \(P + Q = 2\)	Solve for \(P\) (assuming \(a \neq b\))	\(P = 1\)
\(P + Q = 2\)	Substitute \(P=1\), Solve for \(Q\)	\(Q = 1\)
Find \(P - Q\)	Calculate \(1 - 1\)	\(0\)

Solving the System of Equations

Step-by-Step Calculation

Revision Table: System of Equations

Additional Information: Algebraic Manipulation

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