system of equations?

Question

3x + y = 5
4 2

x - y = 1
4 2
============
x = 6
y = 1 I need help in understanding how to find the answers, thanks.

Arrg. This is supposed to be 3x over 4 + y over 2 = 5

x over 4 - y over 2 = 1

Answer 1

3x/4+y/2=5
x/4-y/2=1

Multiply both equations by 4 to get rid of the fractions:
3x+2y=20
x-2y=4
========
4x=24
x=6

6-2y=4
-2y=-2
y=1

The solution set is (6,1).

Check:
3(6)/4+1/2=5
18/4+1/2=5
18/4+2/4=5
20/4=5
5=5

6/4-1/2=1
6/4-2/4=1
4/4=1
1=1

I hope this helps!

Answer 2

If you add up the terms of the two equations, you should obtain an equivalent equation. Just add the terms to the left and the terms to the right of the equal sign:

3x/2 + y/2 +x/4 - y/2 = 6

See how y disappears and we're left with a simple equation where only x is unknown:

3x/4 + x/4 = 6

=> x = 6

Now replace the value of x in one of the other equations to calculate y:

6/4 - y/2 = 1

=> y/2 = 6/4 - 1

=> y/2 = 1/2

=> y=1

The rule of thumb in systems of equations is how to manipulate the equations to be able to get rid of one of the unknown terms, thus reducing the system to a simple equation with only one unknown.

Here's a simple example:
3x + 2y = 7
x + y = 3

Adding these two equations will not reduce any of the unknown terms. However, equivalent equations may be obtained by multiplying all terms of an equation with the same constant value. To reduce y, you would have to multiply the second equation's terms by -2, and obtain
-2x -2y = -6

Now add them up:
3x + 2y -2x -2y = 7 - 6
=> x = 1

Again, replace x=1 in one of the equations to get the value of y.
Simple.

Answer 3

3x + y = 5- - - - - - -Equation 1
x - y = 1- - - - - - - - Equation 2
- - - - - - - - -

Substitute Method equation 2

x - y = 1

x - y + y = y + 1

x = y + 1

Substitute the x value into equation 1

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

3x + y = 5

3(y + 1) + y = 5

3y + 3 + y = 5

4y + 3 = 5

4y + 3 - 3 = 5 - 3

4y = 2

4y / 4 = 2 / 4

y = 2 / 4

y = 1/2

Insert the y value into equation 2

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

x - y = 1

x - 1/2 = 1

x - 1/2 + 1/2 = 1 + 1/2

x = 2/2 + 1/2

x = 3/2

Insert the x value into equation 2

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

Check for equation 2

x - y = 1

3/3 - 1/2 = 1

2/2 = 1

1 = 1

- - - - - - - - - -

Check for equation 1

3x + y = 5

3(3/2) + 1/2 = 5

9/2 + 1/2 = 5

10/2 = 5

5 = 5

- - - - - - - - - - -

Both equations balance

The solution set is { 3/2, 1/2 }

- - - - - - - - - -s-

Answer 4

I presume
3x + y = 5
&
x - y = 1
are your 2 equations
so,we'l add em up

x - y + 3x + y = 5 + 1
=> 4x = 6
=> x = 6/4 = 3/2

putting x = 3/2 in the 2nd equation

3/2 - y = 1
=> - y = 1 - 3/2
Taking 2 as Lowest Common Factor
=> - y = 2 - 3/2
=> - y = -1/2
=> y = 1/2

Answer 5

the start line is to come across a thanks to combine both equations into one ideally eliminating between the unknowns interior the technique. So: x-y=3 might want to be rearranged to x=3+y Substituting this into the first equation provides (3+y)squared + y squared = 29 remedy this to get an answer for y then replace this variety for y interior the 2d equation and remedy to get x. answer: y = 2 x = 5

Answer 6

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

(1) + (2)

(3/4)x + (y/2) + (x/4) - (y/2) = 5 + 1
x = 6

sub x = 6 into (1)

(18/4) + (y/2) = 5
y/2 = 1/2
y = 1