The quotient of 28 divided by 7increased by 4

How do I use elimination process when solving two variable equations?

kenny6666 [7]

Solve simultaneously:

<span>15x-8y=52 and
6x+14y=38
</span>
Let's solve this system thru elimination by addition/subtraction:

Let's get rid of the x terms.
To do this, mult. the first eqn. by -6 and the second by 15. (This is not the only way in which we can do this!!) We get

-90x + 48y = -312
90x +210y = 570
--------------------------
Add all 3 columns together. We end up with:

258y = 258, so y = 1. You can find x by subst. 1 for y in either of the given equations.

If you'd like more help, please be specific about what you want.

7 0

3 years ago

Consider the following three statements:

Lena [83]

It's correlation. A causal relationship between two events exist if the occurrence of the first causes the other. The first event is called the cause and the second event is called the effect. A correlation between two variables does not imply causation. It may sound complicated but think of it like this. They are co related but they don't really effect one another, they just happen to occur at the same time. I really hope this helps. Let me know if I got anything wrong ok!

4 0

4 years ago

Write the expression 7b-3(b-1) in simplest form

Anna007 [38]

4b-3 .. you use distributive property and multiply 3*b-1 then you have 7b-3b-3 then you combine like terms and subtract 3b from 7b and get 4b-3

8 0

3 years ago

Read 2 more answers

What is the missing value?

madam [21]

the answer is 40

in

X 1 (+1) 2 (+1) 3 (+1) 4 (+1) 5

Y 4 (+9) 13 (+9) 22 (+9) 31 (+9) 40

3 0

2 years ago

Read 2 more answers

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago