Question

solve any one question using cross multiplication method and completing of square method.​

Answer 1

Answer:

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Step-by-step explanation:

To find the solution of a pair of linear equations, we use cross multiplication method. If a1x+b1y+c1=0 and a2x+b2x+c2=0 are two linear equations, then we can find the value of x and y using this method.

Answer 2

Answer:

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Questions & Answers

CBSE

Mathematics

Grade 9

Cross-Multiplication Method

Question

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Related Questions

Based on the cross-multiplication method, solve the following pairs of the equation by cross-multiplication rule.

x2−y3+4=0, x2−5y3+12=0x2−y3+4=0, x2−5y3+12=0

Then x + y is equal to

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Hint: We have to only use cross multiplication condition which states that if a1x+b1y+c1=0a1x+b1y+c1=0 and a2x+b2y+c2=0a2x+b2y+c2=0 are the two linear equations then it can also be written as x(b1c2−b2c1)=y(c1a2−c2a1)=1(a1b2−a2b1)x(b1c2−b2c1)=y(c1a2−c2a1)=1(a1b2−a2b1).

Complete step-by-step answer:

To solve the given system of linear equations by cross-multiplication method. We have to first, write them in form of ax+by+c=0ax+by+c=0

Now as we know that we are given with two linear equations and that were,

x2−y3+4=0x2−y3+4=0 →→ (1)

x2−5y3+12=0x2−5y3+12=0 →→ (2)

So, we had to write equation 1 and 2 in the form of ax+by+c=0ax+by+c=0.

So, taking LCM in LHS of equation 1 and equation 2. And then cross multiplying both sides of the equation. We get,

3x – 2y + 24 = 0 →→ (3)

3x – 10y + 72 = 0 →→ (4)

Now as we know that if a1x+b1y+c1=0a1x+b1y+c1=0 and a2x+b2y+c2=0a2x+b2y+c2=0 are the two linear equations then by cross multiplication method. We can write,

x(b1c2−b2c1)=y(c1a2−c2a1)=1(a1b2−a2b1)x(b1c2−b2c1)=y(c1a2−c2a1)=1(a1b2−a2b1)

So, x=(b1c2−b2c1)(a1b2−a2b1)x=(b1c2−b2c1)(a1b2−a2b1) and y=(c1a2−c2a1)(a1b2−a2b1)y=(c1a2−c2a1)(a1b2−a2b1)

So, let us apply the above formula of the two in the equation to find the value of x and y. We get,

x=(b1c2−b2c1)(a1b2−a2b1)=(−2×72−(−10)×24)(3×(−10)−3×(−2))=96−24=−4x=(b1c2−b2c1)(a1b2−a2b1)=(−2×72−(−10)×24)(3×(−10)−3×(−2))=96−24=−4

And, y=(c1a2−c2a1)(a1b2−a2b1)=(24×3−72×3)(3×(−10)−3×(−2))=−144−24=6y=(c1a2−c2a1)(a1b2−a2b1)=(24×3−72×3)(3×(−10)−3×(−2))=−144−24=6

Now x = - 4 and y = 6. So, x + y = 2.