I do not understand how they got the answer to this algebra problem.?

Question

Complete each order pair so that it is a solution of the given linear equation. -4x-5y= -5

( _,1)
(10, _ )

The answer says (0,1) and (10, -7) are the answers but I don't know where they got the -7 from. Can someone explain this?

Answer 1

so they points they gave you: (_,1) and (10,_) are written as coordinates (x,y)
So they're saying for the first one, whatever x you put in, the y is going to be 1:
-4x - 5y = -5
so since we need to find the x, solve for x:
-4x - 5y = -5
-4x = -5 + 5y
x = -5/-4 + 5y/-4
since we already know that y is 1, just plug that in:
x = -5/-4 + 5(1)/-4
x = 5/4 - 5/4 = 0
so x = 0
the point is (0,1)

For the second one we know that x = 10 and are looking for y
so solve for y:
-4x - 5y = -5
-5y = -5 + 4x
y = -5/-5 + 4x/-5
since we know x is 10, substitute that in:
y = -5/-5 + 4(10)/-5
y = 1 - 8 = -7
so y = -7
the point is (10,-7)

Answer 2

The numbers in the parentheses describe the values for x and y. So (0,1) would mean that x = 0 and y = 1. So basically you plug in the values for x and y. When x= 10, the equation becomes -4(10) - 5y = -5, which becomes -40 - 5y = -5. Solve for y.

Answer 3

okay

-4x-5y=-5 rite.
okay so wat u do is u replace the "x" or the "y".

u have
a. (_,1) &
b. (10,_)

in a. u dn't have the "x"
and in b. u dn't have the "y"

u use the linear equation.u 1st replace the "y" or a. with the equation. so, -4x-5(1)=-5.
-4x-5=-5.
u put the -5 on the other side it bcomes positive.
-4x= -5+5.
-4x= 0
x=0. so, u found the "x" of part a.
then, u use the same linear equation for part b.
but u replace the "x" of b.
so, -4(10)-5y= -5
-40-5y= -5
u put the -40 on the other side, then it becomes positive.
-5y= 40-5
-5y= 35.
u divide them both by -5 to get "y"
y= -7.
so, now u found the "y" of b,

conclusion: (0,1) & (10,-7)

Answer 4

Okay, so you have -4x - 5y = -5

Put it in y=mx+b.

-5y = 4x - 5
-5y / -5 = 4x - 5 / -5
y = - 4/5x + 1

So one solution is (0,1) Plug in this answer.

1 = -4/5x + 1 (Correct)

Plug in the other solution. (10, -7)

-7 = -4/5 (10) + 1
-7 = - 8 + 1

Answer 5

Transfer all values/terms to one side of the "equals" then solve the whole algebraic equation -4x - 5y + 5 in the form of ax + bx + c equals 0.

Answer 6

By giving you (10, _ ) they're saying that x is 10 and you need to find y.

So plug x=10 into -4x - 5y = -5 and you get this:
-40 -5y = -5

Now rearrange to get y.

Answer 7

well, the easy way to see where they got it is to graph it.
Then find y on the graph when X is 10.


To double check the answer

-4(10) - 5(-7) = -5
-40 -(-35) = -5
-40 + 35 = -5
-5 = -5

Answer 8

(0,1) and (10,-7) are just coordinates
so it's just plug and play
(x,y)


(_,1):
-4x - 5y = -5
-4x - 5 (1) = -5
-4x - 5 = -5
-4x = 0
x = 0/-4
x = 0

(10,_):
-4x - 5y = -5
-4(10) - 5y = -5
-40 - 5y = -5
-5y = 35
y = 35/-5
y = -7

Answer 9

sounds like a topic for calculating territory (or section). you have L for length and W for width so: length equals Width + 2 length circumstances Width equals the section (Width+2) -which by employing the way is the size- circumstances Witgh equals one hundred twenty meaning that length circumstances Width equals one hundred twenty meaning that the section is one hundred twenty unsure what the little ^ sign means so can no longer assist you to with that section... Width = -12 isn't a valid answer because of the fact width can't be a unfavourable quantity Width equals 10 yards - it is the valid answer for width (that i've got faith you probably did no longer get whilst fixing the priority) if Width is 10 yards, then length = Width + 2 = 12 yards with a view to teach that Width = 10 and Lenght - 12 you are going to be able to desire to multiply them and notice in the event that they equivalent one hundred twenty (the section it is given above) 10*12=one hundred twenty - it is the evidence that length and width have been calculated properly. wish i grow to be of help.

Answer 10

Just substitute the coordinate pair into the equation.
-4*(10)=-40
-5*(-7)= +35
Add them up, -5, q.e.d.