Friday, December 11, 2009

How do you work this example question?

For the systems of linear equations in questions 1-3





- Determine how many solutions exist





- Use either elimination or substitution to find the solutions (if any)





- Graph the two lines, labeling the x-intercepts, y-intercepts, and points of intersection





1. y = 2x + 3 and y = -x - 4


2. 2y + 3x = 7 and 8y + 12x = 28


3. 5x + 6y = 1 and x + 7y = 2How do you work this example question?
QUESTION ONE:





First name your equations ie eQ 1 and eQ 2





eQ 1: y = 2x + 3


eQ 2: y = - x - 4





Using the ';substitution method';, multiply all values of eQ1 by (-1) so that you can eliminate y.





eQ 1: - y = - 2x - 3


eQ 2: y = - x - 4





Add both equations, to eliminate y.





0 = - 3x - 7





Carry all constants on one side of the equation, and all variables to the other side of the equation. Not the sign chages from negative to positive.





7 = - 3x





Multiply both sides of the equation by ( - 1/3) so that x has a new coefficient of positive 1. This is the value of x.





(-1/3)(7) = (-1/3)(-3x)





- 7/3 = x





Subsituting x = - 7/3 into eQ 1: y = 2x + 3





y = 2(-7/3) + 3





y = - 14/3 + 3





y = - 5/3





ANSWER: x = - 7/3 and y = - 5/3





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





QUESTION TWO:





First name your equations:





eQ 1: 2y + 3x = 7


eQ 2: 8y + 12x = 28





Simplify the expressions, and make y the subject of the formula.





eQ 1: 2y = 7 - 3x


eQ 2: 8y = 28 - 12x





eQ 1: (1/2)[ 2y = 7 - 3x ]


eQ 2: (1/8)[ 8y = 28 - 12x]





eQ 1: y = 7/2 - 3/2x


eQ 2: y = 7/2 - 3.2x





Multiply eQ1 by negative 1 so that you can eliminate y.





eQ 1: - y = - 7/2 + 3/2x


eQ 2: y = 7/2 - 3/2x





Based on the above information, there are no solutions for x or y.





ANSWER: N./A





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





QUESTION THREE:





First name your euqations





eQ 1: 5x + 6y = 1


eQ 2: x + 7y = 2





Solve for y, by making it the subject of the formulae





eQ 1: 6y = 1 - 5x


eQ 2: 7y = 2 - x





eQ 1: (1/6)[ 6y = 1 - 5x ]


eQ 2: (1/7)[ 7y = 2 - x ]





eQ 1: y = 1/6 - 5/6x


eQ 2: y = 2/7 - 1/7x





Multiply the entire of eQ 1 by negative 1, so that you can eliminate y.





eQ 1: - y = - 1/6 + 5/6x


eQ 2: y = 2/7 - 1/7x





Add both equations.





0 = 5/42 + 29/42x





Place all constants on one side of the equation, and all variables on the other side.





- 5/42 = 29/42x





Multiply both sides of the equation by (42/29) so that x has a coefficient of 1.





(42/29)(- 5/42) = (42/29)(29/42x)


- 5/29 = x





Subsitute x = - 5/29 into eQ 1: 5x + 6y = 1





5(- 5/29) + 6y = 1





- 25/29 + 6y = 1





6y = 1 + 25/29





6y = 54/29





y = 9/29





ANSWER: x = - 5/29, y = 9/29How do you work this example question?
With a system of 2 linear equations, the maximum number of solutions is 1, because 2 straight lines may cross once or not at all.





1. y = 2x + 3 and y = -x - 4


Using substitution. I substitute the value of y (-x-4) given in the 2nd equation into the first::


(-x-4) = 2x+3 then re-arrange:


-3x = 7


x = -7/3


Now substitute again into either of the equations:


y = 7/3 - 4 = -5/3





Also, I could have used elimination by subtracting one eq from the other:


y = 2x + 3


y = -x - 4


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


0 = 3x + 7


now re-arrange:


3x = -7


x = -7/3 same result as before, which is always nice





now try the others yourself....
  • loan
  • No comments:

    Post a Comment

     
    virus protection