Răspuns:
Explicație pas cu pas:
Metoda substitutiei
e)
{ 2 x - y = 5
{ x + 2 y = 0
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Din ecuatia x + 2 y = 0 => x = - 2 y
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2 x - y = 5
2 × ( - 2 y ) - y = 5
- 4 y - y = 5
- 5 y = 5
y = - 1
x = - 2 y = - 2 × ( - 1 ) ⇒ x = 2
Solutii: ( x; y) = ( 2; - 1 )
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g)
{ 4 x + y = 9 => y = 9 - 4 x
{ - x + 3 y = 1
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- x + 3 y = 1
- x + 3 × ( 9 - 4 x ) = 1
- x + 27 - 12 x = 1
- 13 x = 1 - 27
- 13 x = - 26
x = (-26) : ( - 13 )
x = 2
y = 9 - 4 x = 9 - 4 × 2 = 9 - 8 => y = 1
Solutii: ( x; y ) = ( 2; 1)
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h)
{ x - 2 y + 5 = 0 => { x - 2 y = - 5 => x = - 5 + 2 y
{ 3 x - y - 5 = 0 => { 3x - y = 5
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3 x - y = 5
3 × ( - 5 + 2 y ) - y = 5
- 15 + 6 y - y = 5
5 y = 5 + 15
5 y = 20 => y = 20 : 5 => y = 4
x = - 5 + 2 × 4 = - 5 + 8 => x = 3
Solutii ( x; y) = ( 3; 4)
