If you have 5\6 of a cup of flour, how many twelfths do you have

Solve the following system of equations by elimination. What is the value of x? x = 1.5 x = -1.5 x = 8 x = -8

Andru [333]

Using the elimination method, the value of x in the system of equations is calculated as: 8.

<h3>How to Solve a System of Equations by Elimination?</h3>

To solve a system of equations given using the elimination method, do the following:

Multiply 2x - 5y = 1 by 2 and multiply -3x + 2y = -18 by 5 to get the following:

4x - 10y = 2 --> eqn. 1

-15x + 10y = -90 --> eqn. 2

Add

-11x = -88

Divide both sides by -11

x = 8

Learn more about system of equations on:

brainly.com/question/14323743

#SPJ1

8 0

1 year ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

2 years ago

Which expression gives the measure of

AnnyKZ [126]

What??????????????? lol.

3 0

3 years ago

At the end of the sale, 22 crossbody purses were sold at $15 each, 34 wristlets priced at $12 were sold, 96 satchels marked at $

Brrunno [24]

Answer:

Average: 26.223529411764706 (Please round to the place value necessary since it does not say in the question.)

Step-by-step explanation:

We first find the number of items sold: 170

Then we find the total cost: 4458

We then divide 4458 by 170

We end up with the answer 26.223529411764706. (Please round to the place value necessary since it does not say in the question.)

5 0

2 years ago

Rounding decimals 324,650

Cloud [144]

It depends how many decimal points you are moving. If it's 1 decimal point then the point would move to the right and the answer would be 3246.50if it was two decimal places then it moves two to the right and the answer would 32465.0. But you don't need to put the .0 at the end so it will be 32465

3 0

2 years ago

Read 2 more answers