2x+5y=-55 y=3x+6 What is the solution

Mathematics

2 answers:

zloy xaker [14]3 years ago

6 0

Hey there!!

We have 2 equations.

... 2x+5y=-55 and y=3x+6

Substitution method:

... 2x+5(3x+6)=-55

... 2x+15x+30=-55

... 17x+30=-55

Subtracting 30 on both sides:

... 17x=-55-30

... 17x=-85

Dividing by 17 on both sides:

... x=-85/17

... x = -5

The value of x is -5.

We have:

... 2x+5y=-55

... 2(-5)+5y=-55

... -10+5y=-55

Adding by 10 on both sides:

... 5y=-55+10

... 5y=-45

Dividing by 5 on both sides:

... y=-9

Hence, the value of 'x' is -5 and the value of 'y' is -9.

Hope my answer helps!

Arlecino [84]3 years ago

3 0

Let: eq 1:2x+5y=-55
eq 2:y=3x+6
by substituting eq 2 in eq 1 we get,
2x+5(3x+6)=-55
2x+15x+30+55=0
17x + 85=0
x=-85/17
x=-5
By substituting x value in eq 2,
we have,
y=3(-5)+6=-15+6=-9

You might be interested in

Graph -3x + 9y = -27

Svetllana [295]

Answer:

Step-by-step explanation:

The first step you want to do is get one variable by itself. We are going to start by isolating y by itself.

-3x + 9y = -27

9y = 3x - 27

y = 1/3x - 3

By isolating y by itself, you now have a function written in slope-intercept form. To graph this function, you can draw a point at (0,-3), then use rise/run to go up 1 unit and over 3 units to the next point (3,-2). You can then draw a line based on this information.

8 0

3 years ago

Which polynomial function has a leading coefficient of 1 and roots (7 i) and (5 – i) with multiplicity 1? f(x) = (x 7)(x – i)(x

Nuetrik [128]

It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.

The simplest polynomial with these characteristics would be

$f(x) = (x - (7 + i)) (x - (5 - i))$

but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with

$f(x) = (x - (7 + i)) (x - (7 - i)) (x - (5 - i)) (x - (5 + i))$