Weegy: The answer is y=(x)/(2)+5. [ Here is the steps for solving such a problem: x-2y=6_(-2,4)
Since x does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting x from both sides.
-2y=-x+6
Divide each term in the equation by -2.
-(2y)/(-2)=-(x)/(-2)+(6)/(-2)
Move the minus sign from the denominator to the front of the expression.
-(-(2y)/(2))=-(x)/(-2)+(6)/(-2)
Multiply the factor by the rest of the expression to remove the fraction from the denominator. To multiply by a factor in the denominator, multiply by 1 over the factor.
-*-(2y)/(2)=-(x)/(-2)+(6)/(-2)
Cancel the common factor of 2 in -(2y)/(2).
-*( 2 y)/( 2 )=-(x)/(-2)+(6)/(-2)
Remove the common factors that were cancelled out.
-*-y=-(x)/(-2)+(6)/(-2)
Multiply - by -y to get y.
y=-(x)/(-2)+(6)/(-2)
Move the minus sign from the denominator to the front of the expression.
y=-(-(x)/(2))+(6)/(-2)
Multiply the factor by the rest of the expression to remove the fraction from the denominator. To multiply by a factor in the denominator, multiply by 1 over the factor.
y=-*-(x)/(2)+(6)/(-2)
Multiply - by -x to get x.
y=(x)/(2)+(6)/(-2)
Move the minus sign from the denominator to the front of the expression.
y=(x)/(2)-((6)/(2))
Multiply the factor by the rest of the expression to remove the fraction from the denominator. To multiply by a factor in the denominator, multiply by 1 over the factor.
y=(x)/(2)-(6)/(2)
Cancel the common factor of 2 in -(6)/(2).
y=(x)/(2)-(^(3) 6 )/( 2 )
Remove the common factors that were cancelled out.
y=(x)/(2)-3
To find the slope and y intercept, use the y=mx+b formula where m=slope and b is the y intercept.
y=mx+b
Using the y=mx+b formula, m=(1)/(2).
m=(1)/(2)
To find an equation that is parallel to x-2y=6, the slopes must be equal. ]
(More) 8
y=(1/2)x+6 is the standard form equation of the line that passes through (-2, 4) and is parallel to x - 2y = 6.
Added 10/24/2011 12:27:23 PM
This answer has been confirmed as correct and helpful.
5
How can you rate me bad when the answer is correct. What is would be the standard form...
Added 10/24/2011 8:22:27 PM