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justify your answers:

 

2x + 3y = 20

6x - y = 20

 Aug 3, 2017
 #1
avatar
+1

Solve the following system:
{2 x + 3 y = 20 | (equation 1)
6 x - y = 20 | (equation 2)
Swap equation 1 with equation 2:
{6 x - y = 20 | (equation 1)
2 x + 3 y = 20 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{6 x - y = 20 | (equation 1)
0 x+(10 y)/3 = 40/3 | (equation 2)
Multiply equation 2 by 3/10:
{6 x - y = 20 | (equation 1)
0 x+y = 4 | (equation 2)
Add equation 2 to equation 1:
{6 x+0 y = 24 | (equation 1)
0 x+y = 4 | (equation 2)
Divide equation 1 by 6:
{x+0 y = 4 | (equation 1)
0 x+y = 4 | (equation 2)
Collect results:
Answer: | x = 4        and            y=4
 

 Aug 3, 2017
 #2
avatar+2340 
+1

The question does not ask for the solution of the system of equations but rather how many solutions exist. There are only 3 possibilities. I will list them for you.

 

1. No solution

2. One solution

3. Infinitely many solutions

 

Currently, both equations are written in the form of \(Ax+By=C\). However, in a system, the system looks like the following:
 

\(A_1x+B_1y=C_1\)

{\(A_2x+B_2y=C_2\)

 

Now, let's compare the ratios of the system

 

{\(2x+3y=20\)

{\(6x-y=20\)

 

When I say compare the ratio, I mean that you want to look at the relationship of \(\frac{A_1}{A_2}\) and \(\frac{B_1}{B_2}\) and \(\frac{C_1}{C_2}\).

 

Let's look at them:

 

\(\frac{A_1}{A_2}=\frac{2}{6}=\frac{1}{3}\)

\(\frac{B_1}{B_2}=\frac{3}{-1}=-3\)

\(\frac{C_1}{C_2}=\frac{20}{20}=1\)

 

Since \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), there is only one solution.

 Aug 3, 2017
 #3
avatar+98196 
+2

 

Here's one more method :

 

Using X2's nomenclature

 

If product  of  [A1 * B2 ] - [ A2 * B1 ]   is not 0   then we will always have one solution...so...

 

[2 * -1 ]  - [6 * 3 ] =     -2 - 18   =   -20    .....so....we will have only one solution

 

 

cool cool cool

 Aug 3, 2017
 #4
avatar+2340 
+1

Neat! Very clever, indeed.

TheXSquaredFactor  Aug 3, 2017
 #5
avatar+71 
0

It's just the intersection of two lines.   Since they only intersect at one point,there is only one solution. 

 Aug 3, 2017
 #6
avatar+2340 
+1

To say that 2 nonlinear lines intersects at a common point is correct. However, it is not given info that they are indeed nonlinear; you must prove it.

TheXSquaredFactor  Aug 4, 2017

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