If ab + xy - xb = 0 and bc + yz - cy = 0, then what \(\frac{x}{a} + \frac{c}{z} \) equal to?
1
Understanding the Problem
We are given two algebraic equations involving variables a, b, c, x, y, and z. Our goal is to find the value of the expression \( \frac{x}{a} + \frac{c}{z} \) using these equations.
The given equations are:
- \( ab + xy - xb = 0 \)
- \( bc + yz - cy = 0 \)
We need to manipulate these equations to find expressions for \( \frac{x}{a} \) and \( \frac{c}{z} \) and then add them together.
Solving the First Equation for \( \frac{x}{a} \)
Let's take the first equation:
\( ab + xy - xb = 0 \)
We want to isolate terms involving x and a to eventually get \( \frac{x}{a} \). Let's rearrange the terms to group those with x:
\( xy - xb = -ab \)
Factor out x from the left side:
\( x(y - b) = -ab \)
To get \( \frac{x}{a} \), we can divide both sides by \( a(y - b) \). We assume \( a \neq 0 \) and \( y \neq b \) to avoid division by zero.
\( \frac{x(y - b)}{a(y - b)} = \frac{-ab}{a(y - b)} \)
Simplify both sides:
\( \frac{x}{a} = \frac{-b}{y - b} \)
We can rewrite the right side by multiplying the numerator and denominator by -1:
\( \frac{x}{a} = \frac{-b}{-(b - y)} = \frac{b}{b - y} \)
So, we have an expression for \( \frac{x}{a} \):
\( \frac{x}{a} = \frac{b}{b - y} \quad \text{(Equation 3)} \)
Solving the Second Equation for \( \frac{c}{z} \)
Now, let's take the second equation:
\( bc + yz - cy = 0 \)
We want to isolate terms involving c and z to eventually get \( \frac{c}{z} \). Let's rearrange the terms to group those with c:
\( bc - cy = -yz \)
Factor out c from the left side:
\( c(b - y) = -yz \)
To get \( \frac{c}{z} \), we can divide both sides by \( z(b - y) \). We assume \( z \neq 0 \) and \( b \neq y \) to avoid division by zero.
\( \frac{c(b - y)}{z(b - y)} = \frac{-yz}{z(b - y)} \)
Simplify both sides:
\( \frac{c}{z} = \frac{-y}{b - y} \)
We can rewrite the right side:
\( \frac{c}{z} = \frac{-y}{b - y} = \frac{y}{-(b - y)} = \frac{y}{y - b} \)
So, we have an expression for \( \frac{c}{z} \):
\( \frac{c}{z} = \frac{y}{y - b} \quad \text{(Equation 4)} \)
Adding the Expressions \( \frac{x}{a} \) and \( \frac{c}{z} \)
Now we need to find the sum \( \frac{x}{a} + \frac{c}{z} \). We use the expressions from Equation 3 and Equation 4.
\( \frac{x}{a} + \frac{c}{z} = \frac{b}{b - y} + \frac{y}{y - b} \)
Notice that the denominators are related: \( y - b = -(b - y) \). We can rewrite the second term to have the same denominator as the first term:
\( \frac{y}{y - b} = \frac{y}{-(b - y)} = -\frac{y}{b - y} \)
Substitute this back into the sum:
\( \frac{x}{a} + \frac{c}{z} = \frac{b}{b - y} + \left(-\frac{y}{b - y}\right) \)
\( \frac{x}{a} + \frac{c}{z} = \frac{b}{b - y} - \frac{y}{b - y} \)
Now that the denominators are the same, we can combine the numerators:
\( \frac{x}{a} + \frac{c}{z} = \frac{b - y}{b - y} \)
Assuming \( b - y \neq 0 \), the expression \( \frac{b - y}{b - y} \) simplifies to 1.
\( \frac{x}{a} + \frac{c}{z} = 1 \)
Conclusion
Based on the given equations and assuming the necessary variables are non-zero (specifically \( a \neq 0, z \neq 0, b \neq y \)), the value of \( \frac{x}{a} + \frac{c}{z} \) is 1.
| Step | Action | Result |
|---|---|---|
| 1 | Rearrange first equation: \(ab + xy - xb = 0\) | \(x(y - b) = -ab\) |
| 2 | Solve for \( \frac{x}{a} \) | \( \frac{x}{a} = \frac{b}{b - y} \) |
| 3 | Rearrange second equation: \(bc + yz - cy = 0\) | \(c(b - y) = -yz\) |
| 4 | Solve for \( \frac{c}{z} \) | \( \frac{c}{z} = \frac{y}{y - b} \) |
| 5 | Add \( \frac{x}{a} \) and \( \frac{c}{z} \) | \( \frac{x}{a} + \frac{c}{z} = \frac{b}{b - y} + \frac{y}{y - b} \) |
| 6 | Simplify the sum | \( \frac{b - y}{b - y} = 1 \) |
Revision Table: Key Steps in Solving
This table summarizes the main steps taken to solve the problem involving the algebraic equations and find the value of the expression.
| Concept Used | Application |
|---|---|
| Rearranging Equations | Grouping terms with specific variables (x, c). |
| Factoring | Pulling out common variables (x, c). |
| Division | Dividing both sides to isolate desired ratios \( \frac{x}{a} \) and \( \frac{c}{z} \). |
| Combining Fractions | Adding fractions with related denominators. |
| Simplification | Reducing the final expression to its simplest value. |
Additional Information: Algebraic Manipulation Tips
Solving problems like this often involves basic algebraic techniques. Here are a few tips:
- Isolating Variables: To solve for a specific variable or expression, use inverse operations (addition/subtraction, multiplication/division) to move other terms to the opposite side of the equation.
- Factoring: Look for common factors in terms to simplify expressions, as seen when factoring out x and c in the given equations.
- Combining Fractions: To add or subtract fractions, they must have a common denominator. You can rewrite fractions by multiplying the numerator and denominator by the same non-zero value. Remember that \( a - b = -(b - a) \).
- Checking for Restrictions: When dividing, always consider what values would make the denominator zero. In this problem, we assumed \( a \neq 0, z \neq 0 \), and \( b - y \neq 0 \) (i.e., \( b \neq y \)). If \( b = y \), the original equations simplify differently, possibly leading to conditions like \( ab = 0 \) and \( bz = 0 \), which might make the expression \( \frac{x}{a} + \frac{c}{z} \) undefined or require a different approach depending on the values of a and z. However, standard problems like this usually imply the generic case where denominators are non-zero.
Similar Questions
If (x 2- 1) is a factor of ax 4+ bx 3+ cx 2+ dx + e, then which one of the following is correct?
If α, β and γ are the zeros of the polynomial f(x) = ax 3+ bx 2+ cx + d, then α 2+ β 2+ γ 2is equal to
Let f(x) and g(x) be two polynomials (with real coefficients) having degree 3 and 4 respectively. What is the degree of f(x) g(x)
If α and β are the two zeros of the polynomial 25x 2– 15x + 2, then what is a quadratic polynomial whose zeros are (2α) -1 and (2β) -1 ?
The expression \(\frac{{\left( {{x^3} - 1} \right)\left( {{x^2} - 9x + 14} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - 8x + 7} \right)}}\) simplifies to
If \(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{b-a}\) , then which one of the following is correct?
Which one of the following is a factor of the polynomial
(x - 1)(x - 2)(x - 4) - 90 ?
Consider the following statements in respect of the polynomial a(b - c) (x - b) (x - c) + b(c - a) (x - c) (x - a) + c(a - b) (x - a) (x - b):
1. The coefficient of x2 is 0.
2. The coefficient of x is (a - b) (b - c) (c - a).
Which of the statements given above is/are correct ?
What are the factors of x 3+ 4x 2– 11x – 30?
If 5x 3+ 5x 2– 6x + 9 is divided by (x + 3), then the remainder is
Important Questions from Polynomials
If y 2= y + 7, then what is the value of y 3?
Factorize x 2- y 2- 9z 2+ 6yz
If one of the zeros of the polynomial x 3+ ax 2+ bx + c is - 1, then the product of other two zeros is equal to :
If a(a + b + c) 2 = 1792; b(a + b + c) 2 = 1536; c(a + b + c) 2 = 768, then what will be the value of b?
If x = 3 so, what is the value of x 2 + 2x + 5 ?
