solve algebraically: 3x + 7y = 5; 6x + y = 5? explain please.?

solving the second equation for y:
y = 5 - 6x

Plug this into the first equation
3x + 7(5 - 6x) = 5
3x + 35 - 42x = 5
-39x = -30
x = 30/39
x = 10/13

Plug this into the second equation to get y
6(10/13) + y = 5
y = 5 - 60/13
y = 5/13

3x+7y=5 ...(1)
6x+y=5 ....(2)

Multiply (1) by 2;

6x+14y=10 ...(3)

Subtract (2) from (3);

13y=5
y=5/13

x= (5-y)/6 = 10/13

Ans: Solution set { x=10/13 , y=5/13 }

3x+7y=5 --------------eq(1)

6x+y=5 ----------------eq(2)

2X(1)-eq(2) implies 13y=5

y=5/13

6x+5/13=5[substituting in eq(2)]

6x=5-5/13=60/13

x=10/13

3x + 7y = 5
7y = 5 - 3x
y = (5 - 3x)/7

6x + y = 5
6x + (5 - 3x)/7 = 5
7[6x + (5 - 3x)/7] = 7(5)
42x + 5 - 3x = 35
42x - 3x = 35 - 5
39x = 30
x = 30/39
x = 10/13

3x + 7y = 5
30/13 + 7y = 5
7y = 5 - 30/13
7y = 65/13 - 30/13
7y = 35/13
y = (35/13)/7
y = 35/13 x 1/7
y = 35/91
y = 5/13

we multiply (-2) in first equation>>> -6x-14y=-10
now we have :
-6x-14y= -10
6x+y=5
now we plus this two equation:
-6x+6x-14y+y= -10+5>>>-13y= -5>>>y=5/13

now for x we put y in one of equations:
6x+5/13 =5>>>6x=60/13>>>x=10/13

result:{x=10/13 , y=5/13}

[3x + 7y = 5]---------multiply by 2 to get
6x+14y=10---------eq1
6x + y = 5-----------eq2
subtract eq2 from eq1 to get
13y=5
y=5/13
x=10/13

3x+7y=5(1);6x+y=5(2)
Mul (2) by 7 we get 42x+7y=35(3)
then 3-1=39x=30
x=30/39=10/13
put x=10/13 in (2) we get
6*10/13+y =5
60/13+y=5
y=5-60/13
y=5/13 (x,y)=(10/13,5/13)

3x+7y=5;6x+y=5

the way to do these is to get one integer (x or y) by itself on one side, (even if the answer has the opposite integer on the other side)

the easiest way to do this is to take the second part,

6x+y=5 and solve for y,

so subtract 6x from both sides and you get

y=5-6x

now you take the value of y (which is now 5-6x) and plug it in to the first part of the problem, 3x+7y=5

so wherever there is a y value we plug in (5-6x)

like this

3x+7(5-6x)=5

then solve for x

distribute,

3x+35-42x=5

combine like terms,

-39x=-30

divide -30 by -39

x=1.3

now you take that value of x and plug it into one of the original problems to find y, lets do 6x+y=5

plug in the x value

6(1.3)+y=5

multiply

7.8+y=5

subtract

y=-2.8

(1.3,2.8)

YAY!

I'm pretty sure this is right, but i didn't have a calculator so check the math!

I hope this helped.

The one that (she) used is called the subtitution method...

For me... that is a very difficult to use...

I will give you an easier method.... the elimination method:

In order to do this, coefficients (the number before the letter) of the same variable must be same or opposite...

In that case:

3x + 7y = 5
6x + y = 5

Notice that you can multiply the first equation to 2 to make:

6x + 14y = 10
6x + y = 5

Since there are same coefficients of x... (6x and 6x)

You can subtract the two equations to eliminate x...

6x + 14y = 10
-(6x + y = 5)
--------------------------

13y = 5
y = 5/13

To solve for x, you may subtitute the value of y (5/13) to either of the two equations... But it is better to subtitute that to equation 2 (but subtituting to eq. 1 will also work)

To equation 2:

6x + (5/13) = 5

Multiply the whole equation to 13 to avoid working with fractions...

78x + 5 = 65
78x = 65 - 5
78x = 60
x = 60/78 or 30/39

I will try to equation 1....

3x + 7(5/13) = 5

3x + 35/13 = 5

39x + 35 = 65

39x = 65-35

39x = 30

x = 30/39 (you'll also get the same)

Pls mark me best answer....

3x + 7y = 5
- 42x - 7y = - 35----ADD

- 39x = - 30
x = 30 / 39
x = 10 / 13

60 / 13 + y = 5
y = 65/13 - 60/13
y = 5 / 13

x = 10 / 13 , y = 5 / 13