Plane A travels at an average speed of $x$ km/h for 3 hours 20 minutes and then at an average speed of $y$ km/h for 1 hour 10 minutes. The plane travels a total distance of 3700 km.

1. Convert the travel times into hours and form the distance equation.

$3$ hours $20$ minutes $=3+\frac{20}{60}=\frac{10}{3}$ hours

$1$ hour $10$ minutes $=1+\frac{10}{60}=\frac{7}{6}$ hours

Using $\text{distance}=\text{speed}\times\text{time}$,
$\frac{10}{3}x+\frac{7}{6}y=3700$

2. Clear fractions to show the required simplified equation.

Multiply both sides by $6$:
$6\left(\frac{10}{3}x+\frac{7}{6}y\right)=6(3700)$
$20x+7y=22200$

1. Convert Plane B’s travel times to hours and form the distance equation.

Plane B travels $350$ km less than Plane A, so its total distance is
$3700-350=3350\text{ km}.$

$2$ hours $30$ minutes $=2\frac{30}{60}=\frac{5}{2}$ hours.

$1$ hour $50$ minutes $=1\frac{50}{60}=\frac{11}{6}$ hours.

Using $\text{distance}=\text{speed}\times\text{time}$,
$\frac{5}{2}x+\frac{11}{6}y=3350.$

1. Write the two equations in a form suitable for elimination.

From part (a),
$20x+7y=22200 \quad \cdots (1)$
From part (b),
$\frac{5}{2}x+\frac{11}{6}y=3350$
Multiply by $6$:
$15x+11y=20100 \quad \cdots (2)$

2. Eliminate one variable and solve for $x$.

Multiply $(1)$ by $11$:
$220x+77y=244200 \quad \cdots (3)$
Multiply $(2)$ by $7$:
$105x+77y=140700 \quad \cdots (4)$
Subtract $(4)$ from $(3)$:
$115x=103500$
$x=900$

3. Substitute $x=900$ into one equation to find $y$.

Substitute $x=900$ into $(1)$:
$20(900)+7y=22200$
$18000+7y=22200$
$7y=4200$
$y=600$