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# Is (10, 2) a solution to this system of equations? y = 5x 2x + y = 14

i don't know this answer can you help me?

### 2 Answers by Expert Tutors

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Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Hailey;
y = 5x
2x + y = 14
Let's take the second equation and isolate y...
2x + y = 14
Let's subtract 2x from both sides...
-2x+2x+y=-2x+14
SECOND EQUATION...y=-2x+14
FIRST EQUATION.......y=5x
Let's subtract the first equation from the second...
0=-7x+14
Let's add 7x to both sides...
7x+0=7x-7x+14
7x=14
x=2
Let's plug-in x=2 into the first equation..
y=5x
y=(5)(2)
y=10
Let's plug-in x=2 into the second equation to verify y=10...
y=-2x+14
10=[(-2)(2)]+14
10=-4+14
10=10

The two lines meet at (2,10).
Your question asked if these lines meet at (10,2) thus rendering such as a "solution".
The answer is NO.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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If (10, 2) is a solution to this system of equations:
y = 5x
2x + y = 14

it will make both equations true. So we just perform a check:

y = 5x: 2 =? 5 * 10 => NO

No need to check other equation.

### Comments

Hi Steve;
I appreciate your answer.  However, I noticed immediately that while (10,2) did NOT work, (2,10) did.  Hailey may have copied it incorrectly, or maybe she does not understand that the correct order is (x,y).  I always keep in mind the likely scenario that the instructor wrote it on the board, then the student quickly copied it, then had to read his/her notes a few hours or few days later, and then re-copied it herein.  What we receive may be very different from the original.  We must reconstruct as best we can.
I understand your position; but textbooks often ask students to check if a particular ordered pair is a solution to a system of equations. E.g., Blitzer Precalculus, 2nd Ed, pg 688, Example 1: Determine Whether an Ordered Pair Is a Solution of a System. The Solution given is to substitute the values in both equations to see if both are true.