can you help me to solve th following simultaneous equation2x - 5y = 13 , 7x - 3y = 60?

Question

can you help m to solve the following simulataneous equation
2x - 5y = 13, 7x - 3y = 60
If you could give explanation's as you go that would be a great help

Answer 1

If you multiply each term of the first equation by 7 and the second by 2 the first term will be 14x in each. If you subtract one from the other the x will be gone. Now solve for y. Once you know y you can solve for x.

Answer 2

Here you go:

2x - 5y = 13, x3
7x - 3y = 60 x5

You will times the top eqaution by 3, and the bottom by 5,
this is because the y has to be the same. So now both equations
will have 15 y, and now they will get cancelled out while doing the
equation that will leave you to find out the value of x.

Subtract
6x - 15y= 39
35x - 15y =300


29x = 261

x= 261/ 29

x= 9

Once you got x you replace it in anyone of the equations to get y

I will use the second one

7x - 3y = 60
Since 7x means 7 times x, I replace x with 9, so it will be 7 time 9 which is 63
=63 - 3y = 60

You will send the 63 over to the right side infront of the 60. And since this number
os +63, when you switch it to the other side it will become -63

So now it will be
3y= 60- 63
3y=3
y=3/3
y=1

x= 9 y= 1


To solve the equation you have to either make the x or the y the same number. always make y the same, and get x then use the value of x to get y. You have to see how many times you cane times the equation by to get bot of them the same.

Thank you, and please use that method to solve your further questions

Answer 3

right this is what you do: enable's seem on the equations mutually. 5x + y = - 7 7x - 3y = - 40 5 right this is the deal...you are able to not resolve an equation that has 2 variables except you are able to do away with between the variables. i visit take the 1st equation and multiply all of it the way by way of 3. this might reason the midsection y term to develop right into a 3y...you will see why in a 2nd. 3(5x + y = - 7) 7x - 3y = - 40 5 Now distibute the three... 15x + 3y = - 21 7x - 3y = - 40 5 you are able to now see that in case you combine the two equations, the "y" term would be eradicated. (that's why this gadget is termed the removal approach.) So enable's combine all like words (the 15x with the 7x, the -21 with the -40 5). 22x + 0 = - sixty six Now i think of you are able to resolve the equation and tell me what the fee of x is. 22x = - sixty six x = - 3 YAY!!! yet wait...there is extra...we nevertheless ought to discover the fee of y. Now which you have x, merely "plug" it into any of the unique equations to discover y. 5x + y = -7 5(-3) + y = - 7 - 15 + y = - 7 ...upload 15 on the two aspects of the equivalent sign... y = 8 YAY!!! Now you have your answer...(-3, 8).

Answer 4

cant solve with 2 unknowns (x and y). So eliminate one among them. multiply first with 3, second with 5. you get
3*2x-3*5y=3*13 and 5*7x-5*3y=5*60. Now we have -15y in both equation. So if we substract one equation from another, can eliminate y! and get (3*2x-3*5y=3*13 )-(5*7x-5*3y=5*60) =>29x=261. So x=261/29=9. So if substitute 9 for x in one of the original equation and get 2*9-5*y=13. y is 1.

If this was a part of your homework, please do not try this again. Also please understand the steps here rather than blindly copying it. i might have made mistakes. Math is not tough...a little effort would make you realize that...it is fun actually...

Answer 5

Ok, there are two methods to this. Subsitution and elmination.

Subsitution: your first goal is to set one of the equations equal to a letter. It doesn't matter which one you choose.. we're going to go with "y" in the second equation.

7x-3y=60
-3y=-7x+60
y=7/3x-20

So now that you know what "y" equals, you can put it in your other equation.

2x-5y=13

If y=7/3x-20, then....

2x-5(7/3x-20)=13

(you now solve this equation to find "x")

2x-35/3x+100=13
-29/3x+100=13
-29/3x=-87
x=9

Now that you know that x=9, you can put it into any of the equations to solve for "y".

If x=9, then..

2x-5y=13
2(9)-5y=13
18-5y=13
-5y=-5
y=1

Now you know that y=1, and x=9. Problem solved...

Don't be afraid of fractions.
If you want to know the other method, or if you need any more help, just email me...

Answer 6

Isolate a variable in one of the equations, then substitute for it in the other:
2x-5y=13---->
x=(13+5y)/2 Now substitute for x in equation (2)
7(13+5y/2)-3y=60

Answer 7

Have a look at this person's previous questions! Now I see why education is in such a horrible state. Kiddo PLEASE at least try it yourself. Simultaneuous equations can be solved EASILY by multiplying each equation by the number that x is multiplied by in the other (in this case, multiply the first eqtn by 7 and the second by 2) - this makes each equation have the same number of exes and you can subtract one from the other. Don't forget to multiply the WHOLE equation inlcuding the 'y number' and the total, and that subtracting a negative number is the same as adding a positive one.

But if you were actually going to bother trying this, I guess you would have already. I waste my time.

Answer 8

2x-5y=13.....(1)
7x-3y=60.....(2)
6x-15y=39.....(1) times 3
35x-15y=300...(2) times 5
>>>29x=261 >>>x=9 ........subtraction
sub into (1) >>>>>>y=1

i hope that this helps

Answer 9

6x-15y=39
35x-15y=300 so 29x=261 so x=9 and 54-15y=39 so 15y=15
so y=1

Answer 10

Thisi s the answer for both: x=9 y=1
ma dad dun it lool xxx