Systems in Two Variables: Additional Practice (Set 2)

Parallel lines need matching slopes, so match the coefficient ratios: $\frac{1}{3} = \frac{2}{k}$ gives $k = 6$. Check the constants: the first equation times $3$ is $3x + 6y = 15$, but the second has $8 \ne 15$.

$k = 6,\quad 8 \ne 15 \Rightarrow \text{parallel, no solution}$

For the same line, the first equation must be twice the second: $2(x + 2y = 3)$ is $2x + 4y = 6$.

$k = 4$

Then both equations describe one line, so there are infinitely many solutions.

The second equation is just twice the first, so it says nothing new; both describe the line $x + y = 10$.

$2x + 2y = 20 \ \xrightarrow{\ \div 2\ }\ x + y = 10$

Every pair on that line works, so there are infinitely many.

All three ratios equal means the second equation is a constant multiple of the first: the same line.

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \text{same line}$

So the system is consistent and dependent, with infinitely many solutions.

Exactly one solution fails only when the slopes match. Matching the coefficient ratios: $\frac{k}{2} = \frac{3}{1}$.

$k = 6$

At $k = 6$ the first equation is $6x + 3y = 9$, i.e. $2x + y = 3$, parallel to $2x + y = 4$. Every other $k$ gives exactly one solution.

From the second equation, $y = -2x$. Substitute into the first: $7x + 3(-2x) = 1$.

$x = 1,\quad y = -2(1) = -2$

The solution is $(1, -2)$.

For the same line, the $x$-coefficients must match, since the $y$-coefficients and constants already agree.

$a = 2$

Then both equations are $2x + y = 3$, giving infinitely many solutions.

The second equation is twice the first, so both plot the same line.

$2x + 4y = 8 \ \xrightarrow{\ \div 2\ }\ x + 2y = 4$

You see one line, not two.

The lines share a slope, so they meet only if they are the same line, which needs the second to be twice the first: $2(x - y = 4)$ is $2x - 2y = 8$.

$c = 8 \Rightarrow \text{same line (infinitely many)}$

Every other $c$ gives parallel lines with no solution, so only $c = 8$ has solutions.

The second left side is $-2$ times the first, so the lines coincide only when the whole second equation is $-2$ times the first: $-2(x - 3y = 5)$ is $-2x + 6y = -10$.

$k = -10 \Rightarrow \text{same line (consistent)}$

Any other $k$ gives parallel lines with no solution.

Clear fractions in the first equation by multiplying by $6$: $3x + 2y = 24$. Substitute $x = y + 3$.

$3(y + 3) + 2y = 24 \Rightarrow 5y = 15 \Rightarrow y = 3,\ x = 6$

The solution is $(6, 3)$.

Both lines pass through $(0, 3)$. If the slopes differ they cross only there (one solution); more than one solution needs them to be the same line, i.e. equal slopes.

$m = 2 \Rightarrow \text{same line, infinitely many}$

Only $m = 2$ gives more than one solution.