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2 years ago

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Which first step for solving the given system using substitution results in an equation without fractions? Solve for x in the fi

rst equation. Solve for y in the first equation. Solve for x in the second equation. Solve for y in the second equation.

2 answers:

AfilCa [17]2 years ago

7 0

Answer:

b

Step-by-step explanation:

edit:

the one after that is b

A= 1

B= 2

C= below

lianna [129]2 years ago

5 0

Answer:

solve for x in the second equation is correct

Step-by-step explanation:

Here is the complete question

Which first step for solving the given system using substitution results in an equation without fractions?

(2x+6y=7

(6x + 18y = 24.... 2

Solve for x in the first equation,

Solve for y in the first equation,

Solve for x in the second equation

The best equation to she will be equation 2 since all the coefficients are end numbers

Let us make x t subject f the formula from equation 2;

From 2: 6x+18y = 24

Subtract 18y from both sides

6x+18y-18y = 24-18y

6x = 24-18y

Divide through by 6

6x/6 = 24/6 - 18y/6

x = 4-3y.

You can see that the resulting expression of x is not a fraction. Hence solve for x in the second equation is correct

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2 2/3 multiplied by 5 4/5

grin007 [14]

$2 \frac{2}{3} = \frac{8}{3} 
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$5 \frac{4}{5} = \frac{29}{5}$
$\frac{8}{3} * \frac{29}{5} =X$
Solve for X. What do you think the answer is based on that?

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2 years ago

What is the answer to this problem 75=3(-6n-5)

devlian [24]

Answer:

n=-5

Step-by-step explanation:

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2 years ago

sum of the age of son and his father is60 and the six year ago the age of father is five time the son what is the age son ?

natali 33 [55]

Step-by-step explanation:

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6 0

2 years ago

There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club

ivolga24 [154]

Answer:

The system of equations are $x+y=64$ and $x=y+10$

Step-by-step explanation:

Given : There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club.

To find : Write a system of linear equations that represents the situation.

Solution :

Let x represent the number of students in the drama club

and y represent the number of students in the yearbook club.

There are a total of 64 students in a drama club and a yearbook club.

i.e. $x+y=64$ ....(1)

The drama club has 10 more students than the yearbook club.

i.e. $x=y+10$ ....(2)

Substitute the value of (2) in (1),

$y+10+y=64$

$2y=54$

$y=\frac{54}{2}$

$y=27$

Substitute in (2),

$x=27+10$

$x=37$

Therefore, the system of equations are $x+y=64$ and $x=y+10$

8 0

2 years ago

PLEASE HELP QWQ AsAp with these 4 questions

den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

$\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]$ and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

$(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)]$ which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

$\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right]$ and multiplying gives us

$(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)]$ which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

$\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right]$ and multiplying gives us

$(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)]$ which gives us:

$-126+48-90-[-280+54-36]$ which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

$\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right]$ $\left[\begin{array}{c}x\\y\\\end{array}\right]$ = $\left[\begin{array}{c}-5\\-2\\\end{array}\right]$ and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

$\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]$ (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

$\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right]$ Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

$\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right]$ and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

y = -11

(It works, I checked it)

7 0

2 years ago