Question

common solution for 2x + 7y = -38 and -5x - 8y = 38

Answer 1

Using Elimination Method....Sove For x and y...

Answer 2

2x + 7y = -38
-5x - 8y = 38

We can add the equations together:

2x + 7y = -38
-5x - 8y = 38
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-3x - y = 0

y = -3x

Now put that back into one of the original equations to solve:

2x + 7(-3x) = -38

-19x = -38
x = 2

y = -3x = -3(2) = -6

Answer 3

Can you help me with is too

Answer 4

should i simplify or what?

Answer 5

How do I solve this?

Answer 6

\[(3x+9)/(2x+6) + (8x+12)/(x^2+6x+9)\]

We want a common denominator. Looking at 2x+6 in the denominator of the left-hand fraction, we notice that if we divide it by 2, we get x+3, and (x+3)(x+3) = x^2+3x+3x+9 or x^2+6x+9. Conveniently enough, that's the denominator in the right-hand fraction! So, we multiply top and bottom of the left hand fraction by (1/2)*(x+3) to get a common denominator. As long as we multiply both top and bottom by the same number, we haven't changed the value:

\[(1/2)(x+3)(3x+9)/[(1/2)(x+3)(2x+6)] + (8x+12)/(x^2+6x+9)\] \[= [(1/2)(x+3)(3x+9) + (8x + 12)]/(x^2+6x+9)\]

and if we multiply out the top and collect like terms, we get

\[1/2(3x^2+9x+9x+27) + 8x + 12\] or \[3x^2 + 18x + 27 + 16x + 24\] which gives us \[(3x^2 + 34x + 51)/(x^2+6x+9)\] which you might also write as \[(x(3x+34)+51)/(x+3)^2\]

I hate this sort of tedious, error-prone grunt work, much prefer to have WolframAlpha or similar programs do it for me :-)

Answer 7

:D

Answer 8

By the way, it's sometimes useful to write the denominator in that second form (\[(x+3)^2\]) because it allows you to see the places where the function is going to be ill-behaved. Think about what happens as x approaches -3: you're going to be dividing by a number very close to 0, and the magnitude of the expression will get quite large very rapidly. Some branches of engineering make extensive use of "poles and zeros" to get a feel for how a system will behave. This particular expression has a pole at x = -3 (because the denominator becomes 0) and zeros at values of x that make the numerator equal to 0. We can find those zeros by setting the numerator = 0:
\[x(3x+34) + 51 = 0\]

and using the quadratic formula to find the values of x that make that true.

Answer 9

ohh ok! :)

Answer 10

Old-school engineers know all sorts of tricks for quickly sketching graphs of functions like this, but now everyone uses a computer. It's good to do enough of them by hand that you can look at the computer's answer and decide if it makes sense, or if perhaps you made a mistake typing it in! First rule of computing: Garbage In, Garbage Out :-)

Answer 11

How do i find the distance on the coordinate system from the point (-3, 4) to the point (8, -7).? Am i gonna us the d=(x2-x1)^2 and (y2-y1)^2 formula?

Answer 12

yes, that's how I would do it. the difference in the x values, squared, plus the difference in the y values, squared, take the square root of the sum. if you make the points on a piece of graph paper, you can figure out the sides of a right triangle, and the hypotenuse is the distance between them.