Linear Equations and Systems

1. Plotting the points $(2, 110)$ and $(5, 275)$ on a coordinate plane with time on the $x$-axis and distance on the $y$-axis gives a straight line through the origin.

The slope is:

$m = \frac{275 - 110}{5 - 2} = \frac{165}{3} = 55$

The slope is $55$, which represents the car's constant speed of 55 miles per hour.

2. Runner A's speed is the slope of $d = 6t$, which is $6$ m/s. Runner B's speed is the slope of $d = 5t + 10$, which is $5$ m/s. Runner A is faster.

To find when they are at the same distance, set the equations equal:

$6t = 5t + 10$ $6t - 5t = 10$ $t = 10$

They are at the same distance after $t = 10$ seconds (both at $60$ meters from the start).

3. Yes, this is a proportional relationship because the ratio $\frac{c}{h}$ is constant:

$\frac{12}{1} = \frac{24}{2} = \frac{36}{3} = \frac{48}{4} = 12$

The equation is $c = 12h$.

For 7 hours: $c = 12(7) = 84$. The cost is $84.

4. Add the two equations to eliminate $y$:

$(x + y) + (x - y) = 10 + 4$ $2x = 14$ $x = 7$

Substitute $x = 7$ into the first equation:

$7 + y = 10$ $y = 3$

The solution is $x = 7$, $y = 3$.

Check: $7 + 3 = 10$ and $7 - 3 = 4$.

5. Let $n$ = number of notebooks and $p$ = number of pens.

$n + p = 50$ $3n + p = 90$

Subtract the first equation from the second:

$(3n + p) - (n + p) = 90 - 50$ $2n = 40$ $n = 20$

Substitute into the first equation:

$20 + p = 50$ $p = 30$

The school ordered 20 notebooks and 30 pens.

Check: $20 + 30 = 50$ items, and $3(20) + 1(30) = 60 + 30 = 90$ dollars.