solve linear equations by substitution 3x-5y-4y=0
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(i) We have, 3x−5y−4=0
⇒3x−5y=4...(i)
Again 9x=2y+7
⇒9x−2y=7...(ii)
By Elimination Method:
Multiplying equation (i) by 3, we get
9x−15y=12...(iii)
Subtracting (ii) from (iii), we get
9x−15y=12
9x−2y=7
−13y=5
⇒y=−135
Putting the value of equation (ii), we get
9x−2(−135)=7⇒9x+1310=7⇒9x=7−1310⇒9x=1391−10⇒9x=1381⇒x=139
Hence, the required solution is x=139,
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Answer:
(i) We have, 3x−5y−4=0
⇒3x−5y=4...(i)
Again 9x=2y+7
⇒9x−2y=7...(ii)
By Elimination Method:
Multiplying equation (i) by 3, we get
9x−15y=12...(iii)
Subtracting (ii) from (iii), we get
9x−15y=12
9x−2y=7
−13y=5
⇒y=−135
Putting the value of equation (ii), we get
9x−2(−135)=7⇒9x+1310=7⇒9x=7−1310⇒9x=1391−10⇒9x=1381⇒x=139
Hence, the required solution is x=139,y=−
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