How many solutions does the system have? \begin{cases} y = -2x-4 y = 3x+3 \end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y=−2x−4 y=3x+3 How many solutions does the system have? \begin{cases} y = -2x-4 y = 3x+3 \end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y=−2x−4 y=3x+3 How many solutions does the system have? \begin{cases} y = -2x-4 y = 3x+3 \end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y=−2x−4 y=3x+3

x = -7/5 y = -6/5

Step-by-step explanation:

System:

y = -2x-4

AND

y= 3x+3

therefore -2x-4 = 3x+3

-2x-3x = 3 + 4

-5x = 7

x = -7/5

y = 3(-7/5) + 3 = -21/5 + 3 = -6/5

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.