No Solution or Infinitely Many Solutions

Solution

(a) $y = 2x + 3$ passes through $(-2, -1)$ and $(2, 7)$; $y = 2x - 1$ through $(-2, -5)$ and $(2, 3)$.

For every $x$, the first line is $4$ units above the second. They are parallel and never meet, so no pair $(x, y)$ satisfies both equations.

(b) Divide every term of $2y = 6x + 4$ by $2$: $y = 3x + 2$. It is the same equation as the first, so it gives the same line, through $(-2, -4)$ and $(2, 8)$.

Every point of the line satisfies both equations: $(-1, -1)$, $(0, 2)$, $(1, 5)$ and infinitely many more.

Solution

$x + 2y = 4$ $\quad$ (1)

$2x + 4y = 15$ $\quad$ (2)

(1) $\times 2$: $2x + 4y = 8$ $\quad$ (3)

Subtract (3) from (2): $0x + 0y = 15 - 8$, that is $0 = 7$.

This is false for every $x$ and $y$: no pair can make $2x + 4y$ equal to both $8$ and $15$.

In the form $y = mx + c$: $y = -\frac{1}{2}x + 2$ and $y = -\frac{1}{2}x + 3\frac{3}{4}\text{.}$ Same gradient, different $y$-intercepts: parallel lines.

Solution

$3x - y = 2$ $\quad$ (1)

$6x - 2y = 4$ $\quad$ (2)

(1) $\times 2$: $6x - 2y = 4$, which is (2) exactly. Subtract: $0 = 0$.

This is always true: (2) is (1) multiplied by $2$, so the two equations give the same line, $y = 3x - 2$.

Any value of $x$ gives a solution: $x = 0$, $y = -2$; $x = 1$, $y = 1$; $x = 2$, $y = 4$.

Check $(2, 4)$: $6 - 4 = 2$, true; $12 - 8 = 4$, true.

Solution

(a) $3y = -2x + 6$, so $y = -\frac{2}{3}x + 2$. And $6y = -4x + 5$, so $y = -\frac{2}{3}x + \frac{5}{6}$. Same gradient $-\frac{2}{3}$, different $y$-intercepts ($2$ and $\frac{5}{6}$): parallel.

(b) $y = -\frac{2}{3}x + 2$ and $2y = -3x + 6$, so $y = -\frac{3}{2}x + 3$. Different gradients: the lines meet at one point.

(For (b) the point is $(1.2, 1.2)$: subtracting the equations gives $x - y = 0$, so $x = y$, and $5x = 6$.)

Solution

Multiply every term of $x - 2y = 3$ by $-3$: $-3x + 6y = -9$. This is the second equation, so the two are equivalent and give the same line.

In the form $y = mx + c$, both are $y = \frac{1}{2}x - \frac{3}{2}$: same gradient and same $y$-intercept.

Solutions: make $x$ the subject, $x = 3 + 2y$, and choose $y$. $y = 0$: $(3, 0)$. $y = 1$: $(5, 1)$. $y = -1$: $(1, -1)$. $x = 0$: $-2y = 3$, $y = -1.5$: $(0, -1.5)$.

Check $(5, 1)$ in both: $5 - 2 = 3$; $-15 + 6 = -9$.

Solution

(a) In the second equation the $x$ term $4x$ is $2$ times the $2x$ of the first. For the lines to be parallel, the $y$ term must be $2$ times as well: $6 = 2k$, so $k = 3$.

Then the first equation is $2x + 3y = 5$; doubled, it is $4x + 6y = 10$, but the second says $4x + 6y = 1$. Subtracting gives $0 = 9$: no solution, as wanted.

By gradients: $y = -\frac{2}{k}x + \frac{5}{k}$ and $y = -\frac{2}{3}x + \frac{1}{6}$. Equal gradients need $k = 3$; the intercepts are then $\frac{5}{3}$ and $\frac{1}{6}$, which differ.

(b) Doubling $2x + 3y = 5$ gives $4x + 6y = 10$. For the second equation to be the same line, $k = 10$.

Solution

(a) $x + y = 10$ and $2x + 2y = 25$. Doubling the first gives $2x + 2y = 20$. Subtract: $0 = 5$, which is false.

The facts contradict each other: if the sum is $10$, twice the sum is $20$, not $25$. No such numbers exist.

(b) Let a coffee cost $$x$ and a bun $$y$: $x + y = 3.5$ and $2x + 2y = 7$. The second is the first doubled; subtracting gives $0 = 0$.

The second fact only repeats the first, so the prices cannot be found. Coffee $$2$ and bun $$1.50$ fits, and so do coffee $$1.80$ and bun $$1.70$, and many more.