+0  
 
0
286
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+1965 
+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+86649 
+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+1965 
+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+1965 
+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

10 Online Users

avatar

New Privacy Policy (May 2018)

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see cookie policy and privacy policy.