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monitta
3 years ago
13

Please help me with this!

Mathematics
1 answer:
ivanzaharov [21]3 years ago
4 0

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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Does anyone get this!??
stepladder [879]

Hello from MrBillDoesMath!

Answer:

8

Discussion:

Let the number be "n". Then the question states that

3n - 15 = (n/2) + 5             => multiply both sides by 2

6n - 30 = n + 10                => subtract n from both sides

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Thank you,

MrB

8 0
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