Sent a picture of the solution to the problem (s). I usually graph at least 5 points. Usually one is 0 and I go up from zero and down from zero.
you cant do that because they are not like terms
Answer:
see explanation
Step-by-step explanation:
Given
2BD = 7BT , then
2(d - b ) = 7(t - b) ← distribute both sides
2d - 2b = 7t - 7b ( add 7b to both sides )
2d + 5b = 7t, thus
2![\left[\begin{array}{ccc}10.5\\10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10.5%5C%5C10%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ 5![\left[\begin{array}{ccc}0\\3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C3%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t
![\left[\begin{array}{ccc}21\\20\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ ![\left[\begin{array}{ccc}0\\15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t
![\left[\begin{array}{ccc}21\\35\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C35%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= 7t , thus
t = 
= ![\left[\begin{array}{ccc}3\\5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Hence T = (3, 5 )
3x + 2y = 10 .....multiply by -2
2x + 3y = 15/2...multiply by 3
------------------
-6x - 4y = -20 (result of multiplying by -2)
6x + 9y = 22.5 (result of multiplying by 3)
-----------------add
5y = 2.5
y = 2.5/5
y = 1/2
3x + 2y = 10
3x + 2(1/2) = 10
3x + 1 = 10
3x = 10 - 1
3x = 9
x = 9/3
x = 3
solution is (3,-1/2)
We have 60 gallons of 20% antifreeze.
How much 70% antifreeze do we add to get 60% antifreeze?
We'll make "x" the gallons of 70% we must add.
.20 * 60 + .70 x = .60 * (60 + x)
12 + .70x = 36 + .60x
.10x = 24
x = 240 gallons of 70% antifreeze.
Source
1728.com/mixture.htm (see example B)
