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3 years ago

Sole the following system of equations 8x-3y=13 -3x-8y=-14

Mathematics

1 answer:

user100 [1]3 years ago

8 0

The system reads:

8x-3y=13

-3x-8y=-14

Solve this by solving for one variable at a time by cancelling the other. We will begin by solving for y and cancelling x. To do this, you need to find the least common multiple (LCM) of the x variables and multiply both equations by the number which will make them equal the LCM. For 8 and 3, the LCM is 24, so you will multiply the top equation by 3 and the bottom one by 8.

<em>*****In order for the variables to cancel, the signs need to be opposite. In this system, one is already positive and the other negative; if this were not the case, one of the equations would have needed to be multiplied by a negative number.*****</em>

The system becomes:

24x-9y=39

-24x-24y=42

Now, you can combine the equations by adding them:

-33y=81

Solve for y:

<em>*Divide both sides by -33*</em>

y=27/11

To find x, you can select one of the given equations, plug in 27/11 for y, and solve for x. We'll use the first equation:

8x-3(27/11)=13

8x-81/11=13

<em>*Add 81/11 to both sides*</em>

8x=224/11

<em>*Divide both sides by 8*</em>

x=28/11

Hope this helps!!

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If the demand function for a commodity is given by the equation

Jet001 [13]

Answer:

q=26.92 p=31.13

Step-by-step explanation:

$p^{2} +16q=1400\\\\q=\frac{1400-p^{2} }{16} ;\\\\700-p^{2} +10q=0\\\\q=\frac{p^{2}-700 }{10} ;\\\\$

$q=q\\\\\frac{1400-p^{2} }{16} =\frac{p^{2}-700 }{10} \\\\10(1400-p^{2})=16(p^{2}-700)\\\\14000-10p^{2}=16p^{2}-11200\\\\14000+11200=16p^{2}+10p^{2}\\\\25200=26p^{2}\\\\p^{2}=\frac{25200}{26} \\\\p=+\sqrt{\frac{25200}{26}} \\\\p=31.13$

replace p in any equation and

$q=\frac{1400-p^{2} }{16} \\\\q=\frac{1400-(31.13)^{2} }{16}\\\\q=26.92$

3 0

3 years ago

What is 345,973 rounded to the nearest hundred

8090 [49]

345973 rounded to the nearest hundred is 346000

3 0

3 years ago

Help and please explain so I can do it on my own. There are a couple more on this sheet.

Andreas93 [3]

Answer:

(2, -10)

Step-by-step explanation:

so first i find the axis of symmetry using the equation x= -b/2a

in this problem

a=3

b= -12

c= 2

- (-12)/ 2(3) = 2

2 will be the x point for your vertex

you then plug in your x point into the original equation and solve for y

y = 3(2)^2 - 12(2) +2

y = -10

Vertex is then (2, -10)

8 0

2 years ago

Square root of 32 x square root of 1 over 18

lisov135 [29]

<h3>Solution:</h3>

$\sqrt{32} \times \sqrt{ \frac{1}{18} } \\ = \sqrt{32} \times \frac{ \sqrt{1} }{ \sqrt{18} } \\ = \sqrt{32} \times \frac{1}{ \sqrt{18} } \\ = \frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{ \sqrt{2 \times 3 \times 3} } \\ = \frac{ \sqrt{ {2}^{2} \times {2}^{2} \times 2} }{ \sqrt{2 \times {3}^{2} } } \\ = \frac{2 \times 2 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4}{3}$