$1+2+3+74;$

$5+10+15+50;$

$20+5+30+25; etc.$

$w+x+y+z=80$

$a+b+c=18$

Neebin = a, Clara = b, Gabby = c

$\mathrm{a}=b\times 2$

$c=b\times 3$

So, $a+b+c=18$

$2b+b+3b=18$

$6b=18$

$b=18÷6$

$b=3$

So b is Clara, Clara took 3 pencils.

$a=b\times 2$

$a=3\times 2$

$a=6$

a is Neebin, Neebin took 6 pencils.

$c=b\times 3$

$c=3\times 3$

$c=9$

c is Gabby, Gabby took 9 pencils.

Let Car 1 = a, Car 2 = b, and Car 3 = c.

$a=b+2$

$b=c+4$

1. If car 1 is 30 km, then:

$a=b+2$

$30=b+2$

$b=28$

And

$b=c+4$

$28=c+4$

$c=24$

So, Car 2 is 28km and Car 3 is 24km.

2. If Car 3 is at 25km, then:

$b=c+4$

$b=25+4$

$b=29$

$a=b+2$

$a=29+2$

$a=31$

So, Car 1 is at 31km and Car 2 is at 29km.

3. If the total distance of the three cars is 112km, then:

If $b=c+4$

Then, $c=b-4$

$a+b+c = 112$

$(b+2)+b+(b-4) = 112$

$3b-2 = 112$

$3b = 114$

$b=114÷3$

$b=38km$

• $3\mathrm{m}-7=26\mathrm{;} \mathrm{m}=11$
• $3t=90; t=30$
• $4.2+b-2=20; b=17.8$
• $3y=6.3; y=2.1$
• $1.2g+24.8=56; g=26$
• $x÷6=20.5; x=123$
• $3z-9=66; z=25$