EQ. 1: x + y = 8
EQ. 2: 4x-y = 7
Rewrite EQ. 1 as EQ. 3: x = 8-y
Replace x in EQ.2 with EQ. 3:
4(8-y) - y = 7
Use the distributive property:
32 - 4y - y = 7
Combine like terms:
32 - 5y = 7
Subtract 32 from each side:
-5y = -25
Divide both sides by -5
y = -25 / -5
y = 5
Now replace y with 5 in EQ. 3 to solve for X:
x = 8-5
x = 3
The point of intersection is X = 3, Y = 5, which is written as (3,5)
Answer:
see explanation
Step-by-step explanation:
Given
a = ![\left[\begin{array}{ccc}3\\2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To obtain -3a multiply each of the elements of a by -3
3a = ![\left[\begin{array}{ccc}-3(3)\\-3(2)\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%283%29%5C%5C-3%282%29%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-9\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-9%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To obtain 1.5a multiply each element by 1.5
1.5a = ![\left[\begin{array}{ccc}1.5(3)\\1.5(2)\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1.5%283%29%5C%5C1.5%282%29%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}4.5\\3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4.5%5C%5C3%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The standard form for a line is Ax+By=C
First, we need to find the slope, or change in y over change in x. For the first one, this is
, which is impossible to find as we cannot divide by 0, meaning that this is constant horizontally -- in this case, x=2. Thus, we have 1*x+0*y=2.
For the second one, we can find the slope by getting
. We can then take the point (3,0) (it can be any point on the line) and get our equation to be y-0 = (-2/3) (x-3). Converting this to standard form, we can expand this to get
y= (-2/3)*x +2
(-2/3)*x+1*y = 2
Answer:
The answer is the first one at the tap
