Tracy's Responses in WyzAnt Answers

In this case, it is always good to remember you are distributing a negative one by putting parentheses around it.  For example, you may see other types of problems that won't have the negative in parentheses and it is good practice to do so.

Example:  Add:  2(3x+2) - 4(6x -2)

In this example, we will distribute the 2 across the 3x and +2 to give us 6x +4

Then, we need to see that negative sign in the next operation and ensure it distributes across as a negative 4.  Doing so will give us -24x +8      -4(6x-2) = -24x +8

Now we take 6x +14 -24x +8 and combine like terms to simplify -18x +26

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With the case of -1(4-c), this is a simple version of the the above problem, so we will treat it the same

-1(4-c) ------> -4-(-c) ---------> -4 + c

We could leave it as -4 + c, or flip it around to read c -4

The answer would be 2√6   (read as "2 square root of 6")

Hi -- For this problem, you have already been given some useful information and can use the information to plug in to the standard point-slope formula.  When we talk about point-slope, we mean we have one point defined (-2,-1) and the slope is known (which is 5/1). 

First, we have the point-slope formula:

y-y₁ = m(x-x₁)

Second, we will plug in what we know from what was given in the problem

x₁ = -2  and y₁ = -1

m = 5

Third, let's plug in what we know!

y-(-1) = 5(x-(-2))   Don't forget a (-)(-) = +

y+1 = 5(x+2)         Now, distribute the 5 across (x+2)

y+1 = 5x+10         Now, combine like terms by subtracting 1 from both sides

y = 5x+9              This is the equation of your line with m=5 and y-intercept = (0,9)

This is a system of equations with two variables.   We want to end up with a solution set that looks like this (x, y) where x an y are numerical values.  This is a simple substitution problem where the goal will be to define our variables x & y such that we can then plug them into the original equation and solve for one of them, then repeat the process and solve for the other.

                                              x + 3y = -4  (Call this Equation 1)

                                             4x + 5y = 5   (Call this Equation 2)

In the equations above, we have two variables in two equations.  The variables are x and y.  We can define the value of the variable x in Equation 1 by subtracting 3y from both sides

                                            x + 3y = -4    (Equation 1)

                                               - 3y     -3y

                                            x = -3y - 4     ("new" Equation 1)

By defining the variable x in Equation 1, I can say, "hey look, now I can put the value of x into the second equation anywhere I find an x.

                                               4x + 5y = 5 ( Equation 2)

                                               4(-3y -4) + 5y = 5  (Plug in Equation 1 wherever there is an "x")

                                               -12y -16 + 5y = 5  (distribute the 4 across the binomial)

                                                -7y -16 = 5          (combine like terms)

                                                     +16    +16       (add 16 to both sides)

                                                -7y = 21               (Divide both sides by -7)

                                                    -7

                                                 y = -3

Now, take this value of y = -3 and plug it back into equation 1 to solve for x.

                                               x + 3y = -4

                                               x + 3(-3) = -4

                                               x -9 = -4                 (Add 9 to both sides)

                                                 +9    +9

                                                   x = 5

The solution set for this system of equations with two variables is (5, -3)