It’s 3y not just 3. You need to add the exponent as well
Answer:
y = 5
Step-by-step explanation:
A line parallel to the x- axis has equation
y = c
Where c is the value of the y- coordinates the line passes through
The line passes through (1, 5) with y- coordinate of 5, thus
y = 5 ← equation of line
Y=-3x+5
the slope is negative because the line is going down from left to right
the y-intercept is 5 because that’s where the line crosses the y-axis
A system is inconsistent when there are no solutions between the two equations. Graphically, the lines will be parallel (they never meet!) and the slopes will be the same. But the y-intercepts will be different.
Let's look at the four equations, with each solved as needed, into y = mx + b form.
A: 2x + y = 5
y = 5 - 2x
y = -2x + 5
Compared to y = 2x + 5, the slopes are different, so this system won't be inconsistent. Not a good choice.
B: y = 2x + 5
Compared to y = 2x + 5, the slopes are the same and the y intercepts are the same. This system has infinitely many solutions. Not a good choice.
C: 2x - 4y = 10
-4y = 10 - 2x
-4y = -2x + 10
y = 2/4x -10/4
Here the slopes are different, so, like A this is not a good choice.
D: 2y - 4x = -10
2y = =10 + 4x
2y = 4x - 10
y = 2x - 5
Compared to y = 2x + 5 we have the same slopes and different y intercepts. The lines will be parallel and the system is inconsistent.
Thus, D is the best choice.
Answer:
x= -3 and y= 0
Step-by-step explanation:
5x+2y=-15
<u>2x-2y=-6 </u>
<u>7x =-21</u>
x= -3
Putting value of x in equation 1
5(-3) +2y=-15
-15+2y= -15
2y= 0
y= 0
This can be solved with the help of matrices
In matrix form the above equations can be written in the form
![\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C2%26-2%5C%2F%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let
= A = X and = B
Then AX= B
or X= A⁻¹ B
where A⁻¹= adj A/ ║A║ where mod A≠ 0
adj A= ![\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-2%5C%5C-2%265%5C%2F%5Cend%7Barray%7D%5Cright%5D)
║A║= ( 5*-2- 2*2)= -10-4= -14≠0
X= A⁻¹ B
=- 1/14
=- 1/14 ![\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2A-15%26%2B%20-2%2A-6%5C%5C-2%2A-15%26%2B%205%2A-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2030%26%2B12%5C%5C30%26%2B-30%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc}42\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D42%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-42%2F14%5C%5C0%2F-14%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
From here x= -3 and y= 0
Solution Set = [(-3,0)]
