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# how many solutions does this linear system have?

0
133
6

2x + 3y = 20

6x - y = 20

Guest Aug 3, 2017
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#1
+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
+1221
+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
+76929
+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

CPhill  Aug 3, 2017
#4
+1221
+1

Neat! Very clever, indeed.

TheXSquaredFactor  Aug 3, 2017
#5
+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
+1221
+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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