Systems in Two Variables: Practice

The second left side is twice the first, so the lines coincide exactly when the whole second equation is twice the first: $2(3x - y = 4)$ gives $6x - 2y = 8$.

$k = 2 \cdot 4 = 8$

At $k = 8$ the equations are identical, giving infinitely many solutions.

The second equation is three times the first exactly when $k = 6$: $3(x + 2y = 5)$ is $3x + 6y = 15$.

$k = 6 \Rightarrow 3x + 6y = 15$

Then the equations are the same line, so infinitely many solutions.

Write each total for $v$ visits and set them equal: $20 + 5v = 5v$.

$20 + 5v = 5v \Rightarrow 20 = 0 \text{ (false)}$

The equal-rate lines are parallel, so the costs are never equal; the first plan always costs $20$ dollars more.

The first equation is twice the second, so both are the line $2x - y = 3$. Solve for $y$: $y = 2x - 3$.

$(x,\ 2x - 3)\ \text{for all real } x$

Two lines meet in $0$, $1$, or infinitely many points, never in exactly three.

$\text{possible counts: } 0,\ 1,\ \infty$

So a two-variable linear system cannot have exactly three solutions; the claim is impossible.

Two lines fail to intersect when they are parallel: equal slopes but different intercepts.

$m = 4,\quad 2 \ne -1$

At $m = 4$ both have slope $4$ but different intercepts, so they never meet.

The system $2a + 3c = 37$ and $a + c = 14$ has different slopes, so it meets in one point. Substitute $a = 14 - c$: $2(14 - c) + 3c = 37$.

$28 + c = 37 \Rightarrow c = 9,\ a = 5$

Exactly one price pair works: adult $5$ dollars, child $9$ dollars.

Divide the second equation by $a$: $y = \tfrac{6}{a}x - \tfrac{10}{a}$. Match it to $y = 3x - 5$.

$\frac{6}{a} = 3 \Rightarrow a = 2,\quad \frac{10}{a} = 5 \Rightarrow a = 2$

Both conditions give $a = 2$, so the system is dependent (the same line) there.

Heights after $t$ hours: $A = 10 - 2t$ and $B = 6 - 2t$. Set them equal.

$10 - 2t = 6 - 2t \Rightarrow 10 = 6 \text{ (false)}$

Equal burn rates keep them $4$ cm apart forever, so they are never the same height.

Multiply the first equation by $2$: $x + 2y = 6$, identical to the second.

$0.5x + y = 3 \ \xrightarrow{\ \times 2\ }\ x + 2y = 6$

Same line, so infinitely many solutions.

Both lines have slope $-1$, so they are parallel or identical, never crossing. It is the same line when $k = 6$ (infinitely many) and parallel otherwise (none).

$k = 6:\ \infty \text{ solutions};\quad k \ne 6:\ \text{none}$

Either way it never has exactly one solution.

A single crossing point happens exactly when the lines head in different directions, that is, have different slopes.

$m_1 \ne m_2 \Leftrightarrow \text{exactly one solution}$