Question

The sum of a two-digit number and thenumber obtained by interchanging its digits is 110. If the original number exceeds six times the sum of its digits by 4, find the original number. Solve the equations by the cross-multiplication method.​

Answer 1

Step-by-step explanation:

Given:The sum of a two-digit number and the number obtained by interchanging its digits is 110. If the original number exceeds six times the sum of its digits by 4, find the original number.

To find: Solve the equations by the cross-multiplication method.

Solution:

Cross multiplication method:

If $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are linear equations in two variables, then

using cross multiplication method

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}$

or

$\boxed{\bold{x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}}}\\\\\boxed{\bold{y=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}}}$

Step 1: Write linear equations

Let the number is xy, it can be expressed as 10x+y

the number obtained by interchanging it's digits will be 10y+x

ATQ

$10x + y + 10y + x = 110 \\ \\ 11x + 11y = 110 \\ \\ or \\ \\ x + y = 10 \\ \\ or \\ \\ \bold{\red{ x + y - 10 = 0 }}\: \: \: ...eq1$

If the original number exceeds six times the sum of its digits by 4,

so,

$10x + y - 4= 6(x + y) \\ \\ 10x + y - 4 = 6x + 6y \\ \\ 10x - 6x + y - 6y - 4 = 0 \\ \\ \bold{\green{4x - 5y - 4 = 0}} \: \: \: ...eq2$

Step 2: Apply cross multiplication method

here

$a_1=1,b_1=1,c_1=-10\\\\a_2=4,b_2=-5,c_2=-4$

Put these values in formula

$x=\frac{-4-50}{-5-4} \\ \\ x = \frac{-54}{-9} \\ \\ \bold{x = 6}$

Same way, put the coefficients to find y

$y=\frac{-40+4}{-9}\\\\y = \frac{-36 }{-9} \\ \\\bold{y = 4}$

Final answer:

The two digit number is 64.

Verification:

$64 + 46 = 110$

and

$64 - 4= 6(6 + 4) \\ \\ 60 = 60$

Hope it helps you.

To learn more:

solve by cross multiplication method. x+y=a+b,ax-by=a^2-b^2

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