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Question

# (a)Add: p(p−q),q(q­­­−r) and r(r­−p)(b)Add: 2x(z−x−y) and 2y(z−y−x)(c)Subtract: 3l(l−4m+5n) from 4l(10n−3m+2l)(d)Subtract: 3a(a+b+c)−2b(a−b+c) from 4c(−a+b+c)

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Solution

## (a)First expression =p(p−q)=p2−pqSecond expression =q(q­­­−r)=q2−qrThird expression =r(r­−p)=r2−prAdding the three expressions, we obtain⇒p(p−q)+q(q­­­−r)+r(r­−p)=p2−pq+q2−qr+r2−pr=p2+q2+r2−(pq+qr+rq)Therefore, the sum of the given expressions is p2+q2+r2−pq−qr−rp.(b)First expression =2x(z−x−y)=2xz−2x2−2xySecond expression =2y(z−y−x)=2yz−2y2−2yxAdding the two expressions, we obtain⇒2x(z−x−y)+2y(z−y−x)=2xz−2x2−2xy+2yz−2y2−2yx=2x2−2y2+2xz−2xy−2yx+2yz=2x2−2y2+2xz−4xy+2yzTherefore, the sum of the given expressions is 2x2−2y2+2xz−4xy+2yz(c)First expression =3l(l−4m+5n)=3l2−12lm+15lnSecond expression =4l(10n−3m+2l)=40ln−12lm+8l2Subtracting these expressions, we obtain⇒4l(10n−3m+2l)−3l(l−4m+5n)=40ln−12lm+8l2−(3l2−12lm+15ln)=40ln−12lm+8l2−3l2+12lm−15ln=8l2−3l2+40ln−15ln−12lm+12lm=5l22+25lnTherefore, the result is 5l22+25ln.(d)First expression =3a(a+b+c)−2b(a−b+c)=3a2+3ab+3ac−2ba+2b2−2bc=3a2+2b2+ab+3ac−2bcSecond expression 4c(−a+b+c)=−4ac+4bc+4c2Subtracting these expressions, we obtain,⇒4c(−a+b+c)−[3a(a+b+c)−2b(a−b+c)]=−4ac+4bc+4c2−[3a2+2b2+ab+3ac−2bc]=−4ac+4bc+4c2−3a2−2b2−ab−3ac+2bc=4c2−3a2−2b2−4ac−3ac+2bc+4bc−ab=4c2−3a2−2b2−7ac+6bc−abTherefore, the result is 4c2−3a2−2b2−7ac+6bc−ab.

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