275×0.20 equals $55 withheld. 275-55=220. He brings home $220 each week
X is 7 more than y
x>y then
difference betwen squares is 161 so
x=7+y
and
x²-y²=161
so
x=7+y
sub that for x in other equation
(7+y)²-y²=161
y²+14y+49-y²=161
14y+49=161
minus 49 both sides
14y=112
divide both sides by 14
y=8
sub back
x=7+y
x=7+8
x=15
the numbers are 15 and 8
Answer:
x= -3
y= -2
From the first one, if you add 2y to both sides you can plug that into the second one and solve for y. Then, plug y to the first one and solve for x.
4x + 3y = 54 (1)
3x + 9y = 108 (2)
Multiply (1) by (-3)
-12x - 9y = -162
3x + 9y = 108
---------------------add
-9x = -54
x = 6
plug x = 6 into (1) to find y
4(6) + 3y = 54
24 + 3y = 54
3y = 30
y = 10
Answer
(6 , 10)
Hope it helps.
One way to go about this is to first list everything we know in the form of variables. This will make it easier to see how these numbers correlate instead of trying to remember formulas to plug these numbers into.
TimeA = 2.4h (time of Car A to travel)
TimeB = 4h (time of Car B to travel)
SpeedA = SpeedB + 22mph (Speed of Car A<span>)
</span>SpeedB = SpeedA - 22mph (Speed of Car B<span>)
</span>Distance = x (the distance traveled by each car)
We are looking for SpeedA. How can we find this? Well, we know that speed multiplied by time is equal to distance, so let's start there.
SpeedA * 2.4h = x
<span>(SpeedB + 22mph) * 2.4h = x
</span>(2.4h * SpeedB) + 52.8miles = x
We also know that:
SpeedB * 4h = x
Since both of these equations are equal to x, we can combine them:
SpeedB * 4h = x = <span>(2.4h * SpeedB) + 52.8miles
</span>SpeedB * 4h = <span>(2.4h * SpeedB) + 52.8miles
</span>1.6h * Speed B = 52.8miles
SpeedB = 52.8/1.6 mph = 33 mph
<span>SpeedA = SpeedB + 22mph = 33mph + 22mph = 55mph
</span>
Therefore, Car A was traveling at 55mph.
