Answer:
Step-by-step explanation:
For the graphing, we can graph both equations, and where they intercept is the answer. Okay, so we can give x random inputs and get y as an output which is one pair of coordinates. I usually like using 0 as x. y-3(0)=12. y-0=12. y=12. So for our first equation, on of the coordinates is (0,12). Now we can insert another input for x! ( I chose 1.) y-3(1)=12. y-3=12. y=15. So our other pair of coordinates for the first equation is (1,15). We can do the same with the second equation. 2y+8(0)= -4. 2y+0= -4. y=-2. The first pair of coordinates for the second equation is (0,-2). Another input we can put in is 1, again. 2y+8(1)=-4. 2y+8=-4. 2y= -12. y= -6. So our second pair of coordinates for our second equation is (1,-6). We can graph this with a graphing calculator, or mark these points and draw a straight line through them. When we draw a line through them, the part where the two lines intersect is the answer.
When we do substitution, we need to solve for x or y in the bottom equation. I want to solve for x. ( NOTE: IF YOU SOLVE FOR y, YOU STILL GET THE SAME ANSWER) x=-56-3y. Then we replace the x on the top equation with 56-3y. And we get: 2(56-3y)-y=0. We can use the distributive property. The answer I have is 112-6y-y=0. -6y-y is -7y. 112-7y=0. We can add 7y to both sides so they seperate the variables and the numbers. 112=7y. Lastly, divide by 7. For y, we get 16. To get x, we insert y, AKA 16 into x+3y= -56. x+3(16)=-56. x+48= -56. Our last step to get x is to subtract 48 from both sides leaving us with: x= -100. Our final answer is y= 16 and x= -100.
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Answer:
Somewhere between 12 and 13, perhaps 12.25
Step-by-step explanation:
This is a (rather) simple explanation:
Look for 2 square numbers that are either side of 150
In this case, it is 144 and 169
The square root of 144 is 12 and the square root of 169 is 13
Therefore we can estimate that the square root of 150 is somewhere between 12 and 13.
As 150 is a lot closer to 144 to 169, I would estimate around 12.25 but you do not need an exact value :)
