Finding k for Infinitely Many Solutions of Linear Equations

We are given a system of two linear equations:

1. $kx + 3y = 26$
2. $21x + (k + 2)y = 71 + k$

For a system of linear equations in two variables, $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, to have infinitely many solutions, the ratio of the coefficients must be equal. This condition is given by:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Comparing the given equations with the standard form, we have:

From equation 1: $a_1 = k$, $b_1 = 3$, $c_1 = 26$

From equation 2: $a_2 = 21$, $b_2 = k + 2$, $c_2 = 71 + k$

Now, we apply the condition for infinitely many solutions:

$\frac{k}{21} = \frac{3}{k+2} = \frac{26}{71+k}$

We need to find a value of $k$ that satisfies both equalities. Let's first consider the first two ratios:

$\frac{k}{21} = \frac{3}{k+2}$

Cross-multiplying, we get:

$k(k+2) = 21 \times 3$

$k^2 + 2k = 63$

Rearranging the terms to form a quadratic equation:

$k^2 + 2k - 63 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -63 and add up to 2. These numbers are 9 and -7.

So, the equation can be factored as:

$(k + 9)(k - 7) = 0$

This gives us two possible values for $k$:

• $k + 9 = 0 \implies k = -9$
• $k - 7 = 0 \implies k = 7$

Now, we must check which of these values of $k$ also satisfies the second equality: $\frac{3}{k+2} = \frac{26}{71+k}$.

Checking Possible Values of k

Case 1: $k = 7$

Substitute $k=7$ into the ratios:

• $\frac{k}{21} = \frac{7}{21} = \frac{1}{3}$
• $\frac{3}{k+2} = \frac{3}{7+2} = \frac{3}{9} = \frac{1}{3}$
• $\frac{26}{71+k} = \frac{26}{71+7} = \frac{26}{78}$

To simplify $\frac{26}{78}$, we can divide the numerator and denominator by their greatest common divisor, which is 26:

$\frac{26 \div 26}{78 \div 26} = \frac{1}{3}$

For $k=7$, all three ratios are equal to $\frac{1}{3}$: $\frac{7}{21} = \frac{3}{9} = \frac{26}{78} = \frac{1}{3}$.

Thus, $k=7$ satisfies the condition for infinitely many solutions.

Case 2: $k = -9$

Substitute $k=-9$ into the ratios:

• $\frac{k}{21} = \frac{-9}{21} = -\frac{3}{7}$
• $\frac{3}{k+2} = \frac{3}{-9+2} = \frac{3}{-7} = -\frac{3}{7}$
• $\frac{26}{71+k} = \frac{26}{71+(-9)} = \frac{26}{71-9} = \frac{26}{62}$

To simplify $\frac{26}{62}$, we can divide the numerator and denominator by their greatest common divisor, which is 2:

$\frac{26 \div 2}{62 \div 2} = \frac{13}{31}$

For $k=-9$, the ratios are $-\frac{3}{7}$, $-\frac{3}{7}$, and $\frac{13}{31}$. Since $-\frac{3}{7} \neq \frac{13}{31}$, $k=-9$ does not satisfy the condition $\frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Therefore, the only value of $k$ for which the system of equations has infinitely many solutions is $k = 7$.

This matches one of the given options.

The value of k that yields infinitely many solutions is 7.