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2 years ago

Which ordered pair (a, b) is a solution to the given system? 2a-5b=1 3a+5b=14

Mathematics

1 answer:

irakobra [83]2 years ago

8 0

2a - 5b = 1
3a + 5b = 14

Let's solve this system using elimination. Add the equations together to cancel the b-terms, then solve for a.

2a - 5b = 1
3a + 5b = 14
+___________
5a - 0 = 15
5a = 15
a = 3

Next, plug 3 for a into either of the equations and solve for b.

2a - 5b = 1
2(3) - 5b = 1
6 - 5b = 1
- 5b = -5
b = 1

Lastly, check all work by plugging both values for their respective variables.

2a - 5b = 1
2(3) - 5(1) = 1
6 - 5 = 1 -- This is true

3a + 5b = 14
3(3) + 5(1) = 14
9 + 5 = 14 -- This is true

The solution to the system of equations is (3, 1).

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Multiplying StartFraction 3 Over StartRoot 17 EndRoot minus StartRoot 2 EndRoot EndFraction by which fraction will produce an eq

ziro4ka [17]

The fraction which produce an equivalent fraction with a rational denominator is $\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

Explanation:

The equation is $\frac{3}{\sqrt{17}-\sqrt{2}}$

To find the rational denominator, let us take conjugate of the denominator and multiply the conjugate with both numerator and denominator.

Rewriting the equation, we have,

$\frac{3}{\sqrt{17}-\sqrt{2}}\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

Multiplying, we get,

$\frac{3(\sqrt{17}+\sqrt{2})}{(\sqrt{17})^{2}-(\sqrt{2})^{2}}$

Simplifying the denominator, we get,

$\frac{3(\sqrt{17}+\sqrt{2})}{17-2}$

Subtracting, the values of denominator,

$\frac{3(\sqrt{17}+\sqrt{2})}{15}$

Dividing the numerator and denominator,

$\frac{\sqrt{17}+\sqrt{2}}{5}$

Hence, the denominator has become a rational denominator.

Thus, the fraction which produce an equivalent fraction with a rational denominator is $\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

6 0

2 years ago

Read 2 more answers

You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$