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earnstyle [38]
3 years ago
14

Identify the steps you would take to solve each equation. Then solve 7y + 6 + 4y + 13 = 26​

Mathematics
1 answer:
andriy [413]3 years ago
5 0

Answer:

7/11

Step-by-step explanation:

7y +6+4y+13=26

firstly collect like terms

7y+4y+6+13=26

11y+19=26

11y=26-19

11y=7

then divide through by 11

y=7/11

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Can you help me with this
hichkok12 [17]

Answer:

37/24

Step-by-step explanation:

Since the denomenators must be consistent, both 8 and 12 have the common factor of 24.

the denominator of 8 must be multiplied by 3 to get to 24, but to multiply the denominator you must multiply the numerator by 3 which gives you

15/24

the denominator of 12 must be multiplied by 2 to get 24, so you must multiply 11 by 2 as well to get

22/24

add them together to get 37/24

8 0
3 years ago
The product of 7 and a number, increased by 3, is equal to twice the number, subtracted from 39. Find the number.
musickatia [10]
It's just an equation, the number you're looking for is X, and then turn the words into an solvable equation with respect to X.

7X + 3 = 39 - 2X
9X = 39 -3 = 36
X = 4
8 0
3 years ago
What is the y-coordinate for the solution to the system of equations?<br><br> {y=1/2x−4<br> {y=−2x+1
katovenus [111]

Answer:

y = -3

Step-by-step explanation:

{y = x/2 - 4

y = 1 - 2 x

Substitute y = x/2 - 4 into the second equation:

{y = x/2 - 4

x/2 - 4 = 1 - 2 x

In the second equation, look to solve for x:

{y = x/2 - 4

x/2 - 4 = 1 - 2 x

Add 2 x + 4 to both sides:

{y = x/2 - 4

(5 x)/2 = 5

Multiply both sides by 2/5:

{y = x/2 - 4

x = 2

Substitute x = 2 into the first equation:

{y = -3

x = 2

Collect results in alphabetical order:

Answer:

{x = 2

y = -3

6 0
3 years ago
help please, I have to do it before my class starts its at 10:25 and its 6:21 rn so :) I'm trying to get it done tyyyy
Radda [10]
Ii is the top left and iv is bottom right
7 0
3 years ago
Read 2 more answers
Find the sum of the arithmetic sequence. -1, 2, 5, 8, 11, 14, 17
muminat
56 is you answer you need.
8 0
3 years ago
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