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solve the system {2x+y=8 using substitution method.


{x-y=6 using elimination method


please answer my question ​

Sagot :

Answer:

y = -4/3

x = 14/3

Step-by-step explanation:

2x+y=8

x-y=6

First of all, we need the same coefficient, so we can make both of these 2x by multiplying the second one by 2.

2x+y=8

2x-2y=12

Then we need to subtract equation 1 from 2.

3y=-4

Solve normally

y = -1.33333 / -4/3

Substitute

x - (-4/3) = 6

x = 14/3

[tex]\large\huge\green{\sf{Answer:-}}[/tex]

[tex] \red {\mathbb{ \underline { \tt by \: elimination \: method \: s}}}[/tex]

  • 2x+y=8________(i)
  • x-y=6_________(ii)

from equation (i):-

[tex]2x - y = 8 \\ x = \frac{8 - y}{2} [/tex]

  • put value of x in eq (ii)

[tex]x - y = 6 \\ \frac{8 - y}{2} - y = 6 \\ \frac{8 - y - 2y}{2} = 6 \\ 8 - y - 2y = 12 \\ 8 - 3y = 12 \\ - 3y = 12 - 8 \\ - 3y = 4 \\ y = \frac{ - 4}{3} [/tex]

  • put value of y in eq (i)

[tex]2x + y = 8 \\ 2x + ( \frac{ - 4}{3} ) = 8 \\ 2x - \frac { 4}{3} = 8 \\ \frac{6x - 4}{3} = 8 \\ 6x - 4 = 8 \times 3 \\ 6x - 4 = 24 \\ 6x = 24 + 4 \\ 6x = 28 \\ x = \frac{14}{3} [/tex]

now,

By elimination method:-

[tex]2x + y = 8 \\ x - y = 6 \\ - - - - - - - \\ multiply \: eq \: (i) \: by \: 1 \\ and \: eq(ii) \: by \: 2 \\ \\ (2x + y = 8) \times 1\\ (x - y = 6) \times 2 \\ - - - - - - - \\2x + y = 8 \\ 2x - 2y = 12 \\ subtract \: these \\ 3y = - 4 \\ y = \frac{ - 4}{3} \\ [/tex]

put value of y in equation (i)

[tex]2x + y = 8 \\ 2x + ( \frac{ - 4}{3} ) = 8 \\ 2x - \frac { 4}{3} = 8 \\ \frac{6x - 4}{3} = 8 \\ 6x - 4 = 8 \times 3 \\ 6x - 4 = 24 \\ 6x = 24 + 4 \\ 6x = 28 \\ x = \frac{14}{3} [/tex]

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