Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30

Analysis

We are given a system of two equations:

1. $3x + y = 21$
2. $xy = 30$

We need to find the values of $x$ and $y$ that satisfy both equations simultaneously.

Concept

This is a classic simultaneous equation problem where one equation is linear and the other is non-linear (quadratic). The best approach is the substitution method: express one variable in terms of the other using the linear equation and substitute it into the second equation.

Step-by-Step Solution

Step 1: Express $y$ in terms of $x$ using equation (1). From $3x + y = 21$: $y = 21 - 3x$ --- (Equation 3)

Step 2: Substitute Equation (3) into Equation (2). $x(21 - 3x) = 30$ $21x - 3x^2 = 30$

Step 3: Rearrange into a standard quadratic equation ($ax^2 + bx + c = 0$). Move all terms to one side: $3x^2 - 21x + 30 = 0$

Divide the entire equation by 3 to simplify: $x^2 - 7x + 10 = 0$

Step 4: Factorize the quadratic equation. We look for two numbers that multiply to $10$ and add to $-7$. These numbers are $-2$ and $-5$. $(x - 2)(x - 5) = 0$

So, $x = 2$ or $x = 5$.

Step 5: Find the corresponding values for $y$. Using $y = 21 - 3x$:

• If $x = 2$: $y = 21 - 3(2) = 21 - 6 = 15$
• If $x = 5$: $y = 21 - 3(5) = 21 - 15 = 6$

Thus, the pairs are $(x=2, y=15)$ and $(x=5, y=6)$.

Conclusion

Comparing our results to the options provided: Option C states $x = 2$ or $5$ and $y = 15$ or $6$. This matches our calculated values perfectly.

Exam Tip

When you have a multiple-choice question like this, you can save time by testing the options. Simply plug the values of $x$ and $y$ into the equations. For example, if you test $x=2$ and $y=15$: $3(2) + 15 = 6 + 15 = 21$ (Correct) $2 \times 15 = 30$ (Correct) This confirms Option C is the right choice without needing to solve the full quadratic equation!