Please help Im am Grade 5 and my teacher is letting me do this.

1. Solve the solution by using elimination and substitution

3/x - 2/y = 14
6/x + 3/y = 7

2. Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation

x + y + 5z =2 (1)
4x - 3y + 5z =3 (2)
3x - 2y + 5z=1 (3)

the solution set for this should be {(7,5-2)}

Please I really need your help
teach me how do this

#1 you can solve for 1/x and 1/y in the usual ways:

elimination: double the 1st and subtract
6/x - 4/y = 28
6/x + 3/y = 7
-------------------
7/y = -21
1/y = -3
then, 1/x = 8/3
or, y = -1/3 and x = 3/8

using substitution,
3/x = 14+2/y
2(14+2/y) + 3/y = 7
28 + 4/y + 3/y = 7
7/y = -21
1/y = -3
as above

For elimination, you can enter your coefficients at

http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

and see all the details.

Sure! I'd be happy to help explain how to solve these systems of equations using elimination and substitution.

1. Solve by using elimination and substitution:
To solve this system of equations, we will start by eliminating a variable. Let's eliminate y by multiplying the first equation by 3 and the second equation by 2, so that the coefficients of y will cancel each other out:

3 * (3/x - 2/y = 14) gives us 9/x - 6/y = 42. (4)
2 * (6/x + 3/y = 7) gives us 12/x + 6/y = 14. (5)

Now, add equation (4) and equation (5) together to eliminate y:

(9/x - 6/y) + (12/x + 6/y) = 42 + 14,
21/x = 56.

Next, we can solve for x by multiplying both sides of the equation by x/21:

x/21 * (21/x) = 56 * (x/21),
x = 8.

Now that we have found the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation:

3/x - 2/y = 14.
3/8 - 2/y = 14.

Multiply both sides of the equation by 8y:

(8y * (3/8)) - (8y * (2/y)) = 14 * (8y).

We get:
3y - 16 = 112y.

Combine like terms:
-16 = 112y - 3y,
-16 = 109y,
y = -16/109.

Therefore, the solution to this system of equations is x = 8 and y = -16/109.

2. Solve by eliminating x then substitute into the other two equations:
To solve this system of equations, we will eliminate x by multiplying equation (1) by 4, equation (2) by 1, and equation (3) by 3. This will allow us to cancel out the x's when we add the equations together.

4 * (x + y + 5z = 2) gives us 4x + 4y + 20z = 8. (6)
1 * (4x - 3y + 5z = 3) gives us 4x - 3y + 5z = 3. (7)
3 * (3x - 2y + 5z = 1) gives us 9x - 6y + 15z = 3. (8)

Now, subtract equation (7) from equation (6) to eliminate x:

(4x + 4y + 20z) - (4x - 3y + 5z) = 8 - 3,
7y + 15z = 5. (9)

Now, subtract equation (8) from equation (9) to eliminate x:

(7y + 15z) - (9x - 6y + 15z) = 5 - 3,
13y = 2,
y = 2/13.

Now that we have found the value of y, we can substitute it back into equation (6) to solve for x:

4x + 4(2/13) + 20z = 8,
4x + 8/13 + 20z = 8.

Multiply both sides by 13 to get rid of the fraction:

52x + 8 + 260z = 104.

Combine like terms:

52x + 260z = 96. (10)

Now, substitute y = 2/13 and equation (10) into equation (7) to solve for z:

4x - 3(2/13) + 5z = 3,
4x - 6/13 + 5z = 3.

Multiply both sides by 13 to get rid of the fraction:

52x - 6 + 65z = 39.

Combine like terms:

52x + 65z = 45. (11)

Now, solve equations (10) and (11) simultaneously to find the value of z.

Multiply equation (11) by -5:

-5(52x + 65z) = -5(45),
-260x - 325z = -225.

Now, add equation (10) and the new equation (-260x - 325z = -225) together:

(52x + 260z) + (-260x - 325z) = 96 + (-225),
-65z = -129,
z = 129/65 = 2.

Finally, substitute the values of y = 2/13 and z = 2 into equation (1) to solve for x:

x + 2/13 + 5(2) = 2,
x + 2/13 + 10 = 2,
x + 20/13 = 2,
x = 2 - 20/13,
x = (26 - 20)/13,
x = 6/13.

Therefore, the solution to this system of equations is x = 6/13, y = 2/13, and z = 2.

I hope this explanation helps you understand how to solve systems of equations using elimination and substitution. If you have any more questions, feel free to ask!