- 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
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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
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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
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0 = 3x + 7
now re-arrange:
3x = -7
x = -7/3 same result as before, which is always nice
now try the others yourself....
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