1. Given: 3(m-2) + 14 = 38 Prove: m= 10 please help it’s a proof
2 answers:
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The complete question in the attached figure
we have that
the value of the digit in the hundreds place in the number 653,841 is---> 800
case A) 748,917
the value of the digit in the thousands place------> 8000
1/10 the value of the digit is-------> (1/10)*8000=800
800=800---------------> is ok
case B) 749,817
the value of the digit in the thousands place------> 9000
1/10 the value of the digit is-------> (1/10)*9000=900
900 is not 800
case C) 784,817
the value of the digit in the thousands place------> 4000
1/10 the value of the digit is-------> (1/10)*4000=400
400 is not 800
case D) 797,481
the value of the digit in the thousands place------> 7000
1/10 the value of the digit is-------> (1/10)*7000=700
700 is not 800
the answer is
the option A) 748,917
we have that
the value of the digit in the hundreds place in the number 653,841 is---> 800
case A) 748,917
the value of the digit in the thousands place------> 8000
1/10 the value of the digit is-------> (1/10)*8000=800
800=800---------------> is ok
case B) 749,817
the value of the digit in the thousands place------> 9000
1/10 the value of the digit is-------> (1/10)*9000=900
900 is not 800
case C) 784,817
the value of the digit in the thousands place------> 4000
1/10 the value of the digit is-------> (1/10)*4000=400
400 is not 800
case D) 797,481
the value of the digit in the thousands place------> 7000
1/10 the value of the digit is-------> (1/10)*7000=700
700 is not 800
the answer is
the option A) 748,917
Equation 1: 
Equation 2:
We can either use the 'elimination' method or the 'substitution' method to solve this simultaneous equation.
In some cases one method is easier to use than the other. For this one, it would be easier to use the elimination method
We need to eliminate either the 'x' or the 'y' to begin with. Say we want to eliminate the 'x' terms, then the next step is to make the constant the same
Equation 1: the constant of 'x' is 5
Equation 2: the constant of 'x' is 2
To make the two constants the same, think of common multiple. The lowest common multiple for 5 and 2 is 10
Equation 1: Multiply all terms by 2 to achieve 10x
Equation 2: Multiply all terms by 5 to achieve 10x
Equation 1:
Equation 2:
Once the constant of 'x' is the same, then we can start the elimination process. Since our aim is to eliminate, one of the 'x' need to be in negative form. We can either multiply Equation 1 or Equation 2 by (-1)
Equation 1: -10x - 14y = -62
Equation 2: 10x + 20y = 80
Hence the correct answer is: Third option
Equation 2:
We can either use the 'elimination' method or the 'substitution' method to solve this simultaneous equation.
In some cases one method is easier to use than the other. For this one, it would be easier to use the elimination method
We need to eliminate either the 'x' or the 'y' to begin with. Say we want to eliminate the 'x' terms, then the next step is to make the constant the same
Equation 1: the constant of 'x' is 5
Equation 2: the constant of 'x' is 2
To make the two constants the same, think of common multiple. The lowest common multiple for 5 and 2 is 10
Equation 1: Multiply all terms by 2 to achieve 10x
Equation 2: Multiply all terms by 5 to achieve 10x
Equation 1:
Equation 2:
Once the constant of 'x' is the same, then we can start the elimination process. Since our aim is to eliminate, one of the 'x' need to be in negative form. We can either multiply Equation 1 or Equation 2 by (-1)
Equation 1: -10x - 14y = -62
Equation 2: 10x + 20y = 80
Hence the correct answer is: Third option