For the following system, use the second equation to make a substitution for y in the first equation. 3x + y = 1 y + 4 = 5x what is the resulting equation? 3x + 5x - 4 = 1 3x + 4 - 5x = 1 3x + 5x + 4

3x + y = 1

y + 4 = 5x

Firstly we use the second equation to solve for y , that is we isolate y on left side

y+4 = 5x

subtract 4 from both sides

y = 5x -4

Now we substitute the value of y in first equation

so we get

3x + 5x-4 = 1

If we want we can further solve it for x

8x -4 = 1

add 4 to both sides

8x = 5

divide both sides by 8

x= 5/8

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)