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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)` .
Remove the parentheses around the expression `4-x` .
Since `3x` and `-x` are like terms, add `-x` to `3x` to get `2x` .
Move all terms not containing `x` to the right-hand side of the equation.
Since the equation was solved for `x` , replace all occurrences of `x` in the other equations with the solution `(1)` .
Since the equation was solved for `x` , replace all occurrences of `x ` in the other equations with the solution `(1)` .
Multiply `-1` by each term inside the parentheses.
Subtract 1 from `4` to get `3.`
This is the solution to the system of equations.
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
3x+y=6 - y=-3x+6
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