Pentru sistemele S1, S2 determinati solutiile, utilizand metoda a)substitutiei, b)reducerii
S1:{2x-7y=5
{4x-3y=6
S2:{3x+15y=-5
{2x+2y=6
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Răspuns:
Explicație pas cu pas:
Pentru sistemele S1, S2 determinati solutiile, utilizand metoda a)substitutiei, b)reducerii
S1:{2x-7y=5
{4x-3y=6
S2:{3x+15y=-5
{2x+2y=6
S1: a)
2x-7y=5 ⇒x=(5+7y)/2
4x-3y=6 ⇒4·(5+7y)/2-3y=6 ⇔2·(5+7y)-3y=6 ⇔10+14y-3y=6 ⇔11y=-4 ⇔ y=-4/11
x=(5+7y)/2=[5+7(-4/11)]/2=(5-28/11)/2=(55-28)/22=27/22
x=27/22
S1: b)
2x-7y=5 | ·2 ⇒4x-14y=10
4x-3y=6 4x-3y=6
/ -14y+3y=10-6 ⇔-11y=4 ⇒y=-4/11
2x-7·(-4/11)=5 ⇔2x+28/11=5 ⇔2x=5-28/11 ⇔2x=(55-28)/11 ⇔2x=27/11 ⇒
x=27/22
S2: a)
3x+15y=-5
2x+2y=6 |:2 ⇒x+y=3 ⇒x=3-y
3(3-y)+15y=-5 ⇔9-3y+15y=-5 ⇔12y=-14 ⇔y=-14/12=-7/6 y=-7/6
x=3-y ⇒x=3-(-7/6)=3+7/6=(18+7)/6=25/6 x=25/6
S2: b)
3x+15y=-5 |·2 ⇔6x+30y=-10
2x+2y=6 |·3 ⇔6x+6y=18
/ 30y-6y=-10-18 ⇔24y=-28 ⇔y=-28/24=-7/6 y=-7/6
3x+15·(-7/6)=-5 ⇔3x-35/2=-5 ⇔3x=35/2-5 ⇔3x=(35-10)/2 ⇔3x=25/2 ⇔
x=25/6