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2 years ago

How to solve 4x-3y=22,2x-y=10

2 answers:

katrin2010 [14]2 years ago

8 0

4x-3y=22---->(×1): 4x-3y=22
2x-y=10 ---->(× -2): -4x+2y=-20
_________(+)
0-y=2
y=-2
we replace the value of y in one of the eq
4x-3y=22
4x-3×(-2)=22
4x+6=22
4x=16
x=4
so x=4 and y=-2

Lady bird [3.3K]2 years ago

4 0

Is that considered two equations?

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Y = -4x + 8<br> Y=3/5x - 6/5

VladimirAG [237]

Answer:

x=2 and y=0

Step-by-step explanation:

Rewrite equations:

y=−4x+8;y=

−6

Step: Solvey=−4x+8for y:

y=−4x+8

Step: Substitute−4x+8foryiny=

−6

−4x+8=

−6

−4x+8+

−3

−6

−3

x(Add (-3)/5x to both sides)

−23

x+8=

−6

−23

x+8+−8=

−6

+−8(Add -8 to both sides)

−23

−46

−23

−46

−23

(Divide both sides by (-23)/5)

x=2

Step: Substitute2forxiny=−4x+8:

y=−4x+8

y=(−4)(2)+8

y=0(Simplify both sides of the equation)

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2 years ago

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irina1246 [14]

The value of a is 70° and b is 20°

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2 years ago

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What value of c makes the equation true? −3−3/4(c−4)=5/4 Drag and drop the correct number in the box. c = Response area

gogolik [260]

Answer:

c = -5/3

Step-by-step explanation:

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After you do 11 steps, that is the answer.

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2 years ago

Which direction does the decimal point move in when you divide by powers of 10 greater than 0?

4vir4ik [10]

Answer:

the left side is the answer

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2 years ago

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An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing

wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1 $\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

$S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

$S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

$S=0.9905$

Generally the equation for test statistics is mathematically given by

$t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

$t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$t=-4.00091$

Therefore From table

$P value=P(t_{15+15-2}=4.00091)$

$P value=0.00041918$

$P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

3 0

2 years ago