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3 years ago

Use the elimination method to find all solutions of the system:

Mathematics

1 answer:

weqwewe [10]3 years ago

7 0

Answer:

$x = 3$ and $y = 4$

Step-by-step explanation:

Given

[Correct Question]

$x^2 - 2y = 1$

$x^2 + 5y = 29$

Required

Solve using elimination

First, we eliminate x.

Subtract equation (1) from (2)

$x^2 - x^2 + 5y - (-2y) = 29 - 1\\$

$5y - (-2y) = 28$

$5y +2y = 28$

$7y = 28$

Divide through ny 7

$y = 4$

Take $x^2 - 2y = 1$

Make x^2 the subject

$x^2 = 1 + 2y$

Substitute 4 for y

$x^2 = 1 + 2*4$

$x^2 = 1 + 8$

$x^2 = 9$

Take square roots

$x = \sqrt 9$

$x = 3$

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2 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

5 0

3 years ago