Systems in Two Variables: Additional Practice (Set 1)

Parallel lines need matching slopes: $\frac{2}{1} = \frac{k}{2}$ gives $k = 4$. Check the constants: twice the second is $2x + 4y = 6$, but the first has $5 \ne 6$.

$k = 4,\ 5 \ne 6 \Rightarrow \text{no solution}$

Costs for $g$ GB: $A = 30 + 10g$ and $B = 10 + 10g$. Set them equal.

$30 + 10g = 10 + 10g \Rightarrow 30 = 10 \text{ (false)}$

The equal per-GB rate keeps A exactly $20$ dollars above B, so they never match.

Multiply the first equation by $-2$: $-6x + 8y = -14$, exactly the second.

$-2(3x - 4y = 7) \Rightarrow -6x + 8y = -14$

Same line, so the system is dependent with infinitely many solutions.

Add the equations to eliminate $y$: $(5x - 2y) + (3x + 2y) = 4 + 12$.

$8x = 16 \Rightarrow x = 2,\quad 3(2) + 2y = 12 \Rightarrow y = 3$

The solution is $(2, 3)$.

Inconsistent means parallel: equal slopes with mismatched constants. In $3x + y = 5$ and $6x + 2y = 7$, the second left side is twice the first, but $2 \times 5 = 10 \ne 7$.

$6x + 2y = 10 \ne 7 \Rightarrow \text{parallel}$

So that system has no solution. The others cross once or coincide.

Both equations reduce to $2x + y = 3$, so solve for $y$: $y = 3 - 2x$.

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

Every point on the line is a solution.

The first equation must be twice the second: $2(2x + 3y = 4)$ is $4x + 6y = 8$.

$k = 4$

Then the equations coincide, so there are infinitely many solutions.

The slopes are already both $3$, so the lines coincide only when the intercepts match.

$b = 5$

Then both are $y = 3x + 5$ (infinitely many). Any other $b$ gives parallel lines with no solution.

Rearrange the second equation: $2x - y = 3$ gives $y = 2x - 3$, identical to the first.

$2x - y = 3 \Rightarrow y = 2x - 3$

Same line, so infinitely many solutions.

The system $x + y = 20$ and $x - y = 4$ has different slopes ($-1$ and $1$), so it meets at one point.

$2x = 24 \Rightarrow x = 12,\ y = 8$

Exactly one pair, $12$ and $8$.

No solution means parallel: equal slopes with different intercepts. Making one slope equal the other while keeping the intercepts apart produces exactly that.

$m_1 = m_2,\ b_1 \ne b_2 \Rightarrow \text{no solution}$

Substitute $x = 1$ into $4x + 3y = 10$: $4(1) + 3y = 10$.

$3y = 6 \Rightarrow y = 2$

The solution is $(1, 2)$. A vertical line $x = 1$ still crosses a slanted line exactly once.