Expert: Josh - 9/15/2006

Question
I have 4 questions that I am not understanding how to do. The first one I have been working on for 2 hours and still don't understand it. If you can help me it would be much apprieciated. Please show me the steps if you can.


1. Find the equation, in standard form, with all integer coefficients, of the line perpendicular to 4x – 2y = 10 and passing through (8, 5).

2.  Solve the system of equations using the substitution method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
2x + y = 7
9x + 2y = -1

3. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
6x + 2y = 2
3x + 4y = 1

4. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
4x – 3y = 1
-12x + 9y = -3


Answer
Hi Corey,

Question 1 tests you on the following:
1) You need to know how to read the gradient (aka slope) of a straight line given its equation.
2) Understand what a normal represents.
3) Remember the condition that must be satisfied if a line is to be perpendicular to another line.
4) Recall the general form of an equation for a straight line.

Answers:
1) The equation for a straight line can be written in a number of ways. e.g., y=mx+b is called the gradient form, where "m" represents the gradient (or slope) and "b" represents the y-intercept (the point where the line crosses the y-axis). Of course, if it is written as cx+dy+e=0, we can always rearrange it to y=(-cx-e)/d. In this case, the slope m=-c/d and b=-e/d.

2) A normal is a line perpendicular to another line. i.e., it is oriented at 90 degrees relative to anther line.

3) If we write y=nx+a for the normal to another line y=mx+b, its slope "n" must obey the following condition: n*m=-1. This is very important! Remember this.

4) So, given a straight line y=mx+b, we can work out the gradient of the normal. Specifically, n=-1/m. So, the normal has the form y=nx+a, and we have just determined "n". Now, we are left with one unknown "a" in the normal equation (y=nx+a). To work this out, we generally substitute a known point P=(xo,yo) which lies on the normal. The value of "a" is found by substituting x=xo, y=yo into y=nx+a. Thus, a=yo-n*xo, assuming that a point (xo,yo) was given.

Applying these principles to question 1,
Q1.
1) Given a straight line 4x – 2y = 10, we rearrange it into 2y=4x-10, y=2x-5. Comparing with the standard equation for a straight line, y=mx+b, we have slope m=2, y-intercept b=-5.
3) Normal equation y=nx+a, where n=-1/m=-1/2. Thus, y=-0.5x+a.
4) If P=(xo,yo)=(8,5) lies on the normal, it must satisfy the equation y=-0.5x+a. i.e., 5=-0.5*8+a, a=9.
We have determined that the normal has equation y=-0.5x+9.

The question asks for the normal equation in standard form with integer coefficients. So, we multiply both sides by 2 and rearrange the equation slightly, the steps are shown below.
y=-0.5x+9
2y=-x+18,
x+2y-18=0 is an equivalent representation of the normal. [Some textbook refers to x+2y=18 as the standard form]

Q2.

Strategy: First step using the substitution method is to make one variable (either x, or y -- here we pick y) as the subject of the equation. Then, plug it into the second equation to see if there is an unique solution, infinitely many solutions, or no solution.

Here we go. Rewrite 2x+y=7 as y=-2x+7 ...call this [#1]. Substitute [#1] into second equation 9x+2y=-1.
We obtain 9x+2(-2x+7)=-1.
This is equivalent to
9x+(-4x+14)=-1
5x+14=-1,
5x=-15
x=-3.  Thus, an unique solution exists.
Back substitute this x-value into [#1], we get y=-2*(-3)+7=13.
Do what the question asks for, express the answer as x-y coordinate (x,y)=(-3,13)

Q3. This technique requires you to add/subtract multiples of one equation with another. The idea is to eliminate one of the unknown (either x, or y) in the process, so that you can find the value of (either y, or x).
Let 6x+2y=2 be [#1] and let 3x+4y=1 be [#2].
Observe that in [#1], y has a coefficient of 2. In [#2], y has a coefficient of 4. If we choose to eliminate y first, we can multiply [#1] by -2 to yield -12x-4y=-4...call this [#3].

Now, adding [#3] with [#2], we get
3x+4y + (-12x-4y) = 1+ (-4), cancelling out common terms
-9x = -3
x=1/3.
Again, we can solve this without any problem, so the system of equations has unique solution.

Putting x=-1/3 into [#2] for instance, you can find the solution for y. I'll leave this with you.

Q4. Use the same method as Q3. Except, this time, we notice something strange on the left hand side of the equation.
4x – 3y = 1 ...[#1]
-12x + 9y = -3 ...[#2]
Multiply [#1] by 3 to get 12x -9x =3 ....call this [#3]
Clearly, if we combine [#2] with [#3] by adding the two, we yield a statement -12x+9y+(12x-9y)=-3+3, 0=0 which is unconditionally true. So, there are infinitely many solutions. You can choose any x-value you like, put it into [#1] (or [#2], doesn't matter which one) to obtain a y-value which satisfies the equation. The answer is NOT unique.

Well, it has taken me a while answering these. Hope it helps!

Cheers:)