Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
2. Solving Systems of Linear Equations by Substitution
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Exercise 35 Page 228

47

Practice makes perfect
To find the original number we will need to form and solve a system of equations. To begin, let's define our variables. Let x represent the number in the tens place and let y represent the number in the ones place. Since we know that the sum of these two numbers is 11, we can write our first equation. x+y=11 Now, let's find the second equation. When the digits are kept in their original places, the value of the unknown number can be written as the expression (10x+y). If we swap their positions, we get the expression (10y+x) and the value increases by 27. Let's write this difference as an algebraic equation. (10y+x)-(10x+y)=27 Now that we have two equations, we can solve the system using the Substitution Method.
x+y=11 & (I) (10y+x)-(10x+y)=27 & (II)
x+y=11 10y+x-10x-y=27
x+y=11 9y-9x=27
x+y=11 9(y-x)=27
x+y=11 y-x=3
x+y=11 y=x+3
Now that we have y isolated in the second equation, we can use substitution in the first equation to solve for x.
x+y=11 & (I) y=x+3 & (II)
x+ x+3=11 y=x+3
2x+3=11 y=x+3
2x=8 y=x+3
x=4 y=x+3
We now know that the digit in the tens place is 4. Finally, we substitute this into the second equation to solve for the digit in the ones place.
x=4 & (I) y=x+3 & (II)
x=4 y= 4+3
x=4 y=7
Therefore, the original number was 47.