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earnstyle [38]
3 years ago
7

Solve the system of linear equations 3x-y=10 2x+y=5

Mathematics
1 answer:
omeli [17]3 years ago
7 0

Answer:

The solution for the system of linear equations 3x-y=10  and 2x+y=5 is x=3 y= -1

<u>Solution: </u>

Given that two linear equations are 3x-y = 10 and 2x + y = 5

We have to find the values of “x” and “y”

Let us consider 3x – y =10 ---- eqn 1

2x + y = 5 --- eqn 2

From eqn 1 , rearranging the terms we get

y = 3x-10  --- eqn 3

By substituting the value of “y” from eqn 3 into eqn 2 we get,

2x + 3x – 10 = 5

On solving above expression, 5x – 10 = 5

5x = 15

x = 3

Substitute the value of “x” in eqn 1 to obtain “y” value

3(3) – y = 10

9 – y = 10

y = 9-10 = -1

Hence the solution for the system of linear equations 3x-y=10  

and 2x+y=5 is x=3 and y= -1

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_____________

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Plzz anyone solve all answers plzzzzzzzz​
algol13

You posted a lot of problems here. In the future please only post one problem at a time. Thank you.

I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.

==========================================

Problem 1

{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).

This means we have some number d added on to each term to get the next term.

first term = 18

second term = first term + d = 18+d = a

third term = second term + d = (18+d)+d = 18+2d = b

fourth term = third term + d = (18+2d)+d = 18+3d = -3

----

Let's solve that last equation for d

18+3d = -3

18+3d-18 = -3-18

3d = -21

3d/3 = -21/3

d = -7

----

The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term

first term = 18

second term = first term + d = 18+d = 18+(-7) = 18-7 = 11

third term = second term + d = 11+d = 11+(-7) = 11-7 = 4

fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3

----

We see that a = 11 and b = 4 are the second and third terms respectively.

Therefore, a+b = 11+4 = 15

-------------

<h3>Answer: 15</h3>

==========================================

Problem 2

A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.

We want to know which values of 4*n are between 10 and 250.

----

Divide both 10 and 250 by 4 to get the following

10/4 = 2.5

250/4 = 62.5

If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.

If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62

----

The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"

You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below

{a, a+1, a+2, ..., b-2, b-1, b}

where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula

c = b-a+1

will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}

----

In this case, a = 3 and b = 62, making

c = b-a+1

c = 62-3+1

c = 60

There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}

There are 60 multiples of four that are between 10 and 250.

-------------

<h3>Answer: 60</h3>
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