Since the second equation gives a value for a, we can substitute it into the other equation to find a value for B.
Let's substitute b-2 into the first equation wherever there is an a.
a - 3b = 4
(b-2) - 3b = 4
b - 2 - 3b = 4
-2 - 2b = 4
-2b = 6
b = -3
Now let's find a by substituting -3 into either of the equations to find the value of a.
a = b - 2
a = -3 - 2
a = -5
So your solution set is (-5, -3)
Answer:
In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).
Step-by-step explanation:
There is really no need to use any quadratics or roots.
( Consider the same problem on the plain number line first. )
How do you find the number between 2 and 5 which is twice as far from 2 as from 5?
You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get
4=2+23(5−2)
It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then
R=P+23(Q−P)
so in your case we get
R=(0,−1)+23(3,3)=(2,1)
Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)
Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get
s=a+b=2b+b=3b
⇔b=13s⇒a=23s
Answer:
I believe it's D.
Step-by-step explanation:
x^2 times x
Hope This Helps! Have A Nice Day!!