# y=4-x 3x+y=6

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Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(4-x)` .

`y=4-x `

`3x+(4-x)=6 `

Remove the parentheses around the expression `4-x` .

`y=4-x `

`3x+4-x=6 `

Since `3x` and `-x` are like terms, add `-x` to `3x` to get `2x` .

`y=4-x`

`2x+4=6 `

Move all terms not containing `x` to the right-hand side of the equation.

`y=4-x `

`2x=2 `

Since the equation was solved for `x` , replace all occurrences of `x` in the other equations with the solution `(1)` .

`y=4-x `

`x=1 `

Since the equation was solved for `x` , replace all occurrences of `x ` in the other equations with the solution `(1)` .

`y=4-(1) `

`x=1 `

Multiply `-1` by each term inside the parentheses.

`y=4-1 `

`x=1 `

Subtract 1 from `4` to get `3.`

`y=3 `

`x=1 `

This is the solution to the system of equations.

`y=3 `

`x=1`

Solve the system.

`i. )y = 4 - x`

`ii.)3x + y = 6`

Since equation i. is already solved for y, we will use the substitution method.

Substitute (4-x) for y into equation ii.

`3x + (4 - x) = 6` Solve for x.

`3x + 4 - x = 6`

`2x + 4 = 6`

`2x = 2`

`x = 1` Now that we know that x= 1 then substitute 1 for x into one of the original equations.

`y = 4 - 1`

`y = 3`

**The solution to the system is x = 1 and y = 3 or (1, 3).**

y = 4 - x -------eq(i)

3x + y= 6 -------eq(ii)

These simultaneous equations can be solved with method of substitution, substitute the value of y in eq(ii)

3x + y= 6

3x + 4 - x = 6

2x + 4 = 6

2x + 4 - 4 = 6 - 4 Subtract 4 from both sides

2x = 2

2x/2 = 2/2 Divide both sides by 2

**x = 1**

Now input the value in eq(i)

y = 4 - x

y = 4 - 1

**y = 3 **

**x = 1**

You can check the answer by inserting both values in any equation, if both sides balance the answer is correct.

The set of equations y=4-x and 3x+y=6 has to be solved.

y = 4 - x ...(1)

3x + y = 6 ...(2)

(1) - (2) gives:

y - 3x - y = 4 - x - 6

-3x = -x - 2

Add x to both the sides

-2x = -2

Divide both the sides by -2

x = 1

To determine y using elimination, use 3*(1) - (2)

3y - 3x - y = 12 - 3x - 6

2y = 6

y = 3

The solution of the given set of equations is x = 1 and y = 3

y=4-x y=-x+4

3x+y=6 - y=-3x+6

0=2x-2

2=2x

1=x

y=-(1)+4

y=-1+4

y=3