+0  
 
0
186
6
avatar

justify your answers:

 

2x + 3y = 20

6x - y = 20

Guest Aug 3, 2017
Sort: 

6+0 Answers

 #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
 

Guest Aug 3, 2017
 #2
avatar+1493 
+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.

TheXSquaredFactor  Aug 3, 2017
 #3
avatar+79881 
+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

CPhill  Aug 3, 2017
 #4
avatar+1493 
+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. 

frasinscotland  Aug 3, 2017
 #6
avatar+1493 
+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

9 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details