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`...

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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).**