Solve the system of equations. Please help me

Question

Solve the system of equations:

A) 4w + 5z = 34

z - 2w = 4

w = ?

z = ?

B) r + s = -4

s - 4r = 16

s = ?

r = ?

Answer 1

For each of these systems of equations, part A and B, there are two variables in each part. We are given two equations that relate them to each other. You could say that these equations describe the relationship between the two variables.

One common way to solve a problem like this is to start with one of the variables and try to "isolate" it, or get it alone. If you can isolate one variable, you can use it to find the other. I'll show you how this works. Lets start with part A, and try to isolate one of the variables.

A)

in Part A, we have w and z as our variables. The two equations are:

4w + 5z = 34

z - 2w = 4

Lets start with trying to isolate a variable (getting it by itself). What this means is I would like to adjust or re-express one of these equations to look like "z = _____" or "w=_____".

Looking at our two equations, I think the second will be easier to do this with, I'll show you how:

z - 2w = 4

I'll add 2w to both sides of the equation: z - 2w + 2w = 4 + 2w which simplifies to this:

z = 4 + 2w this is exactly what we want, the variable 'z' isolated by itself on one side of the equals sign. Obviously, we dont know what z is yet, but this new equation we have shows us "z in terms of w". We have a simplified equation telling us what z equals, in terms of w.

Now, we can combine this with our first original equation to find w. Our first equation again is this:

4w + 5z = 34

and we now know that z = 4 + 2w so we can input or substitute that value in for z in the first equation, like this:

4w + 5(4 + 2w) = 34 I subbed in what z equals into that first original equation.

Now, this is a great spot to be in, because we only have a w variable in this equation now. With just one variable, we can simplify solve for the value of that variable w. Like this:

4w + 5(4 + 2w) = 34

4w + 20 + 10w = 34

14w + 20 = 34 I'm going to subtract 20 from both sides now

14w + 20 - 20 = 34 - 20

14w = 14 now divide by 14, to find:

w = 1 so we have w, now lets use it to find z, by plugging in that value of w into one of our equations that has both z and w.

I could choose either of our first equations, or I could choose the adjusted one we made to find z. I'll do the one we made, it was:

z = 4 + 2w and we now know that w = 1, so lets plug in that value for w in our equation:

z = 4 + 2(1)

z = 4 + 2

z = 6 so final answer: w = 1, z = 6

On problems like these, there is a great way to check your work. To make sure you got the right answer, plug in that w value to the original equations. If you plug w = 1 into both of them, you should end up with z = 6 for both of them. If you dont get the same z value from both of those equations, that means somethings wrong and you should go back and double check your work.

B) I'll solve part B with the same method, starting with isolating a variable first then using it to find the other. I'll just show the work below for this one:

r + s = -4

s - 4r = 16 --> + 4r to both sides --> s = 16 + 4r

Now I have s "in terms of r", so I will plug in what s equals into the first equation:

r + s = -4

r + 16 + 4r = -4

5r + 16 = -4 --> - 16 from both sides --> 5r = -20 --> divide both sides by 5 -->

r = -4

Now plug in r = -4 into a different equation to find s. I'll do it to both original equations to see if we get the same answer:

r + s = -4 --> -4 + s = -4 --> s = 0

s - 4r = 16 --> s - 4(-4) = 16 --> s + 16 = 16 --> s = 0

So final answer, r = -4, s = 0

Answer 2

I won't do your homework for you, so demonstrating the first problem should help you solve the second on your own. That is the goal.

4w + 5z = 34

z — 2w = 4

I would choose to solve the above two equations by substitution, meaning that I would manipulate one of the equations so that I have z= or w=. The easier thing to do would be to solve the second equation for z.

z — 2w = 4

Add 2 w to each side of the equation to get: z = 2w + 4 and then substitute z, expressed with w (2w + 4), into the first equation to get:

4w + 5z = 34

4w + 5(2w + 4) = 34

Now do the distribution of multiplication over addition to get:

4w + 10w + 20 = 34

Now combine like terms to get:

14w + 20 = 34

Now isolate the w term on one side of the equation by subtracting 20 from each side to get:

14w = 34 – 20 = 14

14w = 14

Divide through by the coefficient of the w term (14) to get:

w = 1

Now that you know what w is you can substitute 1 for w in either one of the two original equations. I always take the simplest one, so I am going to choose the second one. Here it goes!

z — 2w = 4

z – 2(1) = 4

z – 2 = 4

Add two to each side of the equation to isolate the z term.

z = 4 + 2

z = 6

Your answer: w = 1 and z = 6.

You can check those answers by substituting the values into each of the original two equations and making sure that they each make true statements.

Answer 3

How to solve systems of equations using Elimination and Substitution. For this problem Substitution is the best Method. Fast Forward to about 5 Minutes to see Substitution Method.