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Question

# Solve the following pair of linear equations by the elimination method and the substitution method : (i) x+y=5 and 2x−3y=4 (ii) 3x+4y=10 and 2x−2y=2(iii) 3x−5y−4=0 and 9x=2y+7(iv) x2+2y3=−1 and x−y3=3

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Solution

## (i) x+y=5 ...(i) 2x−3y=4 ...(ii) (elimination method)Multiply eqn (i) by 2 2x+2y=10 2x−3y=4−+−____________ 5y=6⇒y=6/5Substitute y=6/5in eqn (i)x+ y=5x+6/5=5 ⇒x=5×5/5−6/5=19/5 Substitution method x+y=5 ....(i)2x−3y=4 ....(ii)y=5−x ...(iii)Putting eqn. (3) in (2)2x−3(5−x)=42x−15+3x=4x=19/5y=5×5/5−19/5y=6/5(ii)Elimination method 3x+4y=10 ...(i)2x−2y=2 ...(ii)Multiply eqn (2) by 24x−4y=43x+4y=10____________7x=14x=22×2−2y=2⇒4−2=2yy=1Substitution method 3x+4y=10 ...(i)2x−2y=2⇒x−y=1 x−1=y ...(ii)Putting eqn (ii) in (i) 3x+4(x−1)=103x+4x−4=10_________________7x=14x=2y=x−1y=2−1⇒y=1(iii) 3x−5y=4 ...(i)9x−2y=7 ...(ii) (elimination method)Multiply eqn (i) by 3(ii)-(i)×3 9x−2y=7 9x−15y=12 _____________ 13y=−5⇒y=−5/13 3x−5(−5/13)=4⇒3x=4−25/133x=(52−25)/13x=27/(13×3)⇒x=9/13Substitution method 3x−5y=4 ...(i)9x−2y=7 ⇒9x−7=2y(9x−7)/2=y 3x−5y=4⇒3x−5(9x−7)/2=4⇒3x−45/2x+35/2=4 ⇒6x2−452x=4−352 ⇒−39x2=8−352 x=2739 ⇒x=913 Putting x=913 in 3x−5y=4 3×913−5y=4 2713−5y=4 ⇒2713−4=5y⇒27−5213=5y −2513=5y ⇒y=−513 (iv)Elimination methodx2+2y3=−13x+4y6=−1⇒3x+4y=−6 ...(i)x−y3=3⇒3x−y=9 ...(ii) (i)−(ii)3x+4y=−63x−y=−9−+−______________ 5y=−15⇒y=−3Putting y=−3 in eqn (i)3x−12=−6⇒3x=−6+12⇒3x=6 ⇒x=2Substitution method 3x+4y=−6 ...(i)3x−y=9 ...(ii)3x−9=y ....(iii)Putting enq (iii) in (i)3x+4(3x−9)=−6⇒3x+12x−36=−6⇒15x=−6+36⇒15x=30⇒x=2 Putting x=2 in eq (iii)6−9=y⇒y=−3

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