Algebra

Linear Equation

Medium

For some real numbers $a$ and $b$ , the system of equations $x + y = 4$ and $(a + 5)x + (b^2 - 15)y = 8b$ has infinitely many solutions for $x$ and $y$ . Then, the maximum possible value of $ab$ is

$25$

$33$

$55$

$15$

✅ Correct Option: 2

When we have two linear equations in two variables, there are three possibilities:

One unique solution (lines intersect at one point)

No solution (parallel lines that never meet)

Infinitely many solutions (the equations represent the same line)

For infinitely many solutions, the second equation must be a multiple of the first equation.

We are given the system:

Equation 1: $x + y = 4$

Equation 2: $(a + 5)x + (b^2 - 15)y = 8b$

For infinitely many solutions, we need:

$\frac{\text{coefficient of } x \text{ in eq 2}}{\text{coefficient of } x \text{ in eq 1}} = \frac{\text{coefficient of } y \text{ in eq 2}}{\text{coefficient of } y \text{ in eq 1}} = \frac{\text{constant term in eq 2}}{\text{constant term in eq 1}}$

This gives us:

$\frac{a + 5}{1} = \frac{b^2 - 15}{1} = \frac{8b}{4}$

If equation 2 is exactly 2 times equation 1, then $(a+5) = 2 \times 1$ , $(b^2-15) = 2 \times 1$ , and $8b = 2 \times 4$ .

From the proportion: $\frac{b^2 - 15}{1} = \frac{8b}{4}$

$\frac{8b}{4} = 2b$

So: $b^2 - 15 = 2b$

$b^2 - 2b - 15 = 0$

We need two numbers that multiply to $-15$ and add to $-2$ .

These numbers are $-5$ and $+3$

So: $(b - 5)(b + 3) = 0$

Therefore: $b = 5$ or $b = -3$

From the proportion: $\frac{a + 5}{1} = \frac{8b}{4} = 2b$

So: $a + 5 = 2b$

Case 1: When $b = 5$

$a + 5 = 2(5) = 10$

$a = 5$

Case 2: When $b = -3$

$a + 5 = 2(-3) = -6$

$a = -11$

We have two possible pairs:

$(a, b) = (5, 5)$ : $ab = 5 \times 5 = 25$

$(a, b) = (-11, -3)$ : $ab = (-11) \times (-3) = 33$

The product of two negative numbers is positive, and $|-11| \times |-3| = 33 > 25$ .

Therefore, the maximum possible value of $ab$ is $33$ .

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For some real numbers aaa and bbb, the system of equations x+y=4x + y = 4x+y=4 and (a+5)x+(b2−15)y=8b(a + 5)x + (b^2 - 15)y = 8b(a+5)x+(b2−15)y=8b has infinitely many solutions for xxx and yyy. Then, the maximum possible value of ababab is

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For some real numbers $a$ and $b$ , the system of equations $x + y = 4$ and $(a + 5)x + (b^2 - 15)y = 8b$ has infinitely many solutions for $x$ and $y$ . Then, the maximum possible value of $ab$ is