3x+6=6(x+?)-3x ? need help !

I believe you should try to plug in the ordered pair to see if the solutions work.

The ordered pair is (x,y) because x is on the x-axis and y is on the y-axis, therefore in (16,-3) the x=16 and y=-3.

Now substitute, meaning plug in those numbers into the equation.
For x+2y=10 would be 16+2(-3)=10
Now you just need to solve one side and see if it equals the other side.
16+2(-3)
First you use order of operations to solve this. PEMDAS. So you multiply 2(-3) first because of the parenthesis and it being multiplication.
16+(-6) and when you have a positive number adding a negative number it’s going backwards of the number line, basically meaning subtraction in a way. Sorry if this confuses you, if you already know how to do negatives and such nevermind this part.
But 16+(-6)=10
So now you look at both side of the equation, does the left side equal to the right? 10=10, so yes. It is a solution for that equation.

Now for the next equation, 7y=-21
Again, plug in the ordered pair (16,-3) into the equation. Remember that it’s (x,y).
There is no x in this equation so no need to worry about that; you only plug in y for this one.
7(-3) Now you multiply. Whenever you multiply a positive number and a negative number, the answer will always be negative. So 7(-3) is -21.
Now look if the left side is equivalent to the right. Does -21=-21? Yes. The ordered pair is a solution.

(16,-3) is a solution to both equations.

Hope this helps!

5 0

3 years ago

Find the difference quotient off, that is, find fly+h)-f(x) h h#0, for the following function. Be sure to simplify. f(x) = x^2 -

Sloan [31]

The difference quotient for the function f(x) is $h+2x-7$

Given :

Difference quotient formula

$\frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}$

Given function $f(x) = x^2 - 7x + 8$

find the difference quotient using the formula

first we find out f(x+h) using given f(x)

replace x with x+h

$f(x) = x^2 - 7x + 8 \\f(x+h)=\mathrm{ }\left(x+h\right)^2-7\left(x+h\right)+8\\Expand \; and \; simplify\\f(x+h)=x^2+2xh+h^2-7\left(x+h\right)+8\\f(x+h)=x^2+2xh+h^2-7x-7h+8$

Now replace it in our formula and also replace f(x)

$\frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}\\ \frac{x^2+2xh+h^2-7x-7h+8-(x^2 - 7x + 8)}{h} \\\frac{x^2+2xh+h^2-7x-7h+8-x^2 + 7x - 8}{h} \\\\\frac{h^2+2xh-7h}{h}\\factor \; out \; h\\\frac{h\left(h+2x-7\right)}{h}\\h+2x-7$