Question

If \(x y-3 y=2\) and \(x+y=6\), then \(x\) is: A. 4 or 5 B. 3 or 4 C. 4 D. 5 E. Cannot be determined

Short answer

4 or 5

Step-by-step solution

- Rewrite the first equation

From the equation \( x y - 3 y = 2 \), factor out \( y \) to get: \( y(x - 3) = 2 \)

- Rearrange the second equation

From the equation \( x + y = 6 \), express \( y \) in terms of \( x \): \( y = 6 - x \)

- Substitute \( y \) in the first equation

Plug \( y = 6 - x \) into \( y(x - 3) = 2 \): \((6 - x)(x - 3) = 2 \)

- Simplify the equation

Expand and simplify the equation: \( 6x - 18 - x^2 + 3x = 2 \) which simplifies to \( -x^2 + 9x - 18 = 2 \)

- Form a quadratic equation

Rearrange to form a quadratic equation: \( -x^2 + 9x - 20 = 0 \). Multiply through by -1 to simplify: \( x^2 - 9x + 20 = 0 \)

- Solve the quadratic equation

Factorize the quadratic equation: \((x - 4)(x - 5) = 0 \). Thus, \( x = 4 \) or \( x = 5 \)

- Verify the solution

Since we have \( x + y = 6 \), substituting \( x = 4 \) gives \( y = 2 \); substituting \( x = 5 \) gives \( y = 1 \). Both pairs satisfy the original equations.

Key concepts

Understanding Quadratic Equations

A quadratic equation is any equation that can be written in the standard form: \( ax^2 + bx + c = 0 \)where \(a\), \(b\), and \(c\) are constants. These equations are crucial in many areas of mathematics and science because they describe parabolic shapes. In our exercise, we derived the quadratic equation from a system of linear equations by substituting one equation into the other. This process transforms the equations into a single quadratic equation, which we then need to solve.

The Role of System of Equations

A system of equations consists of two or more equations that share common variables. To find solutions to a system, we need to find values for the variables that satisfy all the equations simultaneously. In the given problem, we have:

• \( xy - 3y = 2 \)
• \( x + y = 6 \)

By solving these equations together, we find a common solution for \( x \) and \( y \). The usual approach is either substitution or elimination. Here, we used substitution to replace \( y \) in the first equation with \( 6 - x \) from the second equation.

Factoring Techniques

Factoring is the process of breaking down an equation into simpler expressions that, when multiplied together, give the original equation. For quadratic equations, this often involves finding two binomials whose product equals the quadratic equation. In the exercise, we reached the quadratic equation: \( x^2 - 9x + 20 = 0 \).We factorized it into:\[ (x - 4)(x - 5) = 0 \]Factoring is efficient because it allows us to find the roots of the quadratic equation immediately. Each factor equals zero, giving us the solutions for \( x \):

• \( x - 4 = 0 \), hence \( x = 4 \)
• \( x - 5 = 0 \), hence \( x = 5 \)

Verification Process

Verifying a solution is critical in mathematics, especially when working with systems of equations or quadratic equations. After solving for \( x \) and \( y \), we must substitute these values back into the original equations to ensure they are correct. In our case, substituting \( x = 4 \) and \( x = 5 \) into \( x + y = 6 \) yields:

• For \( x = 4 \), \( y = 6 - 4 = 2 \)
• For \( x = 5 \), \( y = 6 - 5 = 1 \)

Next, we substitute these \( (x, y) \) pairs back into the first equation:

• \( x y - 3 y = 2 \); for \( x = 4 \) and \( y = 2 \):
  • \( 4 * 2 - 3 * 2 = 8 - 6 = 2 \)
  • for \( x = 5 \) and \( y = 1 \) : \( 5 * 1 - 3 * 1 = 5 - 3 = 2 \)

Both pairs satisfy the original equations, confirming our solutions are correct.