Quandaries and Queries
 

 

I hope that u can help me....I am a college student taking a class in Pre Calculus.....I have homework due this Friday and it counts a BIG Percentage on my FINAL grade.....I am getting mixed up and can not figure out a few problems.....Please help me.....

Method Of Subsitution

Problem 1. y- 8x = -5
x(squared) + y(squared) = 25

Problem 2. y = x(squared) - 2x - 6
Y = x(squared) - 4

Problem 3. Solve by elimination
4x - y= -1
3x + 4y = -2

Problem 4. Method of Sub and Method of Elimination
X - 2 Y + Z = 8
3X - 5Y + 10Z = 6
2X - 6Y - 11Z = 4

I hope that u can help me with this....Out of 50 questions, I got stuck on these 4.....Please help ASAP.....

Thanks..
Scott

 

 

Hi Scott,

Since this is such a big homework assignment I am not going to do the problems for you but I will show you solutions to similar problems.

Method of Substitution.

y + 2x = 3
2x2 + y2 = 5 - 2x

Since y + 2x = 3, y = 3 - 2x

Thus, substituting into the second equation

2x2 + (3 - 2x)2 = 5 - 2x
2x2 + 9 -12x + 4x2 = 5 - 2x
6x2 - 10x + 4 = 0
2(3x2 -5x + 2) = 0
2(3x - 2)(x -1) = 0

Thus x = 2/3 or x = 1.

Substitute these values back into he first equation to find the corresponding values of y.

Method of elimination.

4x + 2y = 5
x - 3y = 2

Multiply the second equation by 4. This will make the coefficients of x equal in both equations.

4x + 2y = 5
4x - 12y = 8

Now subtract the second equation from the first. This gives

14y = -3

Thus y = -3/14

Substitute this value back into one of the two equations to find the corresponding value of x.

For both of these problems you should check your answers.

Penny

 
 

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