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2 years ago

Solve each inequality(x+4)(x-8)≥0

Mathematics

1 answer:

Stolb23 [73]2 years ago

8 0

Answer: x

≤

−

≥

Step-by-step explanation:

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Please help me with this!

ivanzaharov [21]

1. The first equation is - 2x + 5y = 0

Second equation is $y = \frac{2}{5} x$

5y = 2x

- 2x + 5y = 0

Hence, the two equations are equivalent.

2. $a_{1} = 2, a_{2} = - 2$

$b_{1} = -1, b_{2} = -1$

$\frac{a_{1} }{a_{2}} =\frac{2}{-2} = -1$

$\frac{b_{1} }{b_{2}} = \frac{-1}{-1} = 1$

$\frac{a_{1} }{a_{2}} \neq \frac{b_{1} }{b_{2}}$

Hence, the equations are consistent.

3. $a_{1} = 4, a_{2} = 6$

$b_{1} = -1, b_{2} = -1$

$\frac{a_{1} }{a_{2}} =\frac{4}{6} = \frac{2 }{3}$

$\frac{b_{1} }{b_{2}} = \frac{-1}{-1} = 1$

$\frac{a_{1} }{a_{2}} \neq \frac{b_{1} }{b_{2}}$

Hence, the equations are consistent.

4. Equations can be re-arranged as:

x + y - 4 = 0 and

x + y + 6 = 0

$a_{1} = 1, a_{2} = 1$

$b_{1} = 1, b_{2} = 1$

$c_{1} = -4, c_{2} = 6$

$\frac{a_{1} }{a_{2}} =\frac{1}{1} = 1$

$\frac{b_{1} }{b_{2}} =\frac{1}{1} = 1$

$\frac{c_{1} }{c_{2}} =\frac{-4}{6} = \frac{-2}{3}$

$\frac{a_{1} }{a_{2}} = \frac{b_{1} }{b_{2}} \neq \frac{c_{1} }{c_{2}}$

Hence, the equations are inconsistent.

5. If we multiply the first equation by 4, we will get,

2y = -4x + 20 which is the second equation.

Hence, the equations are equivalent.

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