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2 years ago

What value of m makes the system of linear equations mx − y + 3 = 0, (2m − 1)x − y + 4 = 0 have no solution?

Mathematics

1 answer:

Morgarella [4.7K]2 years ago

6 0

Answer:

m ≠1 ( all m in R except 1 )

Step-by-step explanation:

hello :

mx − y + 3 = 0.....(*)

(2m − 1)x − y + 4 = 0 ....(**)

multiply (*) by : -1 you have : -mx+y-3=0 ....(***)

(2m − 1)x − y + 4 = 0 ....(**)

add(***) and(**) : -mx+ (2m − 1)x+1 =0

(2m-m-1)x+1=0

(m-1)x = -1

this system have no solution if : m-1≠0 means : m ≠1

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Solve the system of equations by substitution. 3/8 x + 1/3 y =17/24 and <br> x + 7y = 8

goblinko [34]

$\left \{ {{ \frac{3}{8}x+ \frac{1}{3}y = \frac{17}{14} } \atop {x+7y=8}} \right.$

To solve this system by substitution, first isolate x in the second equation.

$x+7y=8$
$x+7y-7y=8-7y$
$x=8-7y$

Now, plug this expression ( $8-7y$ ) for x in the top equation to solve for y.

$\frac{3}{8} (8-7y)+ \frac{1}{3} y= \frac{17}{14}$
$3- \frac{21}{8} y+ \frac{1}{3} y= \frac{17}{24}$
$72-63y+8y=17$
$72-55y=17$
$-55y=17-72$
$-55y=-55$
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Now that you have y, plug it into the second equation and solve for x.

$x+7y=8$
$x+7(1)=8$
$x+7=8$
$x=1$

Last step is to plug your x- and y-values in to both equations to check your work.

$\frac{3}{8} (1)+ \frac{1}{3} y= \frac{17}{24}$

$\frac{3}{8} * \frac{3}{3} = \frac{9}{24} ; \frac{1}{3} * \frac{8}{8} = \frac{8}{24}$

$\frac{9}{24} + \frac{8}{24} = \frac{17}{24}$ <--True

$1+7(1)=8$
$1+7=8$ <--True

Answer:
$x=1 \\ y=1$

5 0

3 years ago

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nordsb [41]

Answer:

$\boxed{a_n \: = \: 64 \: \times \: ( - \frac{3}{4} ) ^{n \: - \: 1} }$

Step-by-step explanation:

We first compute the ratio of this geometric sequence.

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We simplify the fractions:

$r \: = \: - \frac{3 }{4} \\ \\ r \: = \: - \frac{3 }{4} \\ \\ r \: = \: - \frac{3 }{4}$

We deduce that it is the common ratio because it is the same between each pair.

$r \: = \: - \frac{3 }{4}$

We use the first term and the common ratio to describe the equation:

$a_1 \: = \: 64; \: r \: = \: - \frac{3 }{4}$

<h3>We apply the data in this formula:</h3>

$\boxed{a_n \: = \: a_1 \: \times \: {r}^{ n \: - \: 1} }$

_______________________

<h3>We apply:</h3>

$\boxed {\bold{a_n \: = \: 64 \: \times \: {( - \frac{3}{4} )}^{ n \: - \: 1} }}$

<u>Data</u>: The unknown "n" is the term you want

<h3><em><u>MissSpanish</u></em></h3>

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nataly862011 [7]

Answer:

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Step-by-step explanation:

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ahrayia [7]

Answer:

Step-by-step explanation:

The question is incomplete. Here is the complete question.

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