Asked by Unknown
                Solve the following system of equations for a and for b
:
System of Equations:
	
Value of a
	
Value of b
9a+3b=30
		
8a+4b=28
	
	
            
            
        :
System of Equations:
Value of a
Value of b
9a+3b=30
8a+4b=28
Answers
                                                    There are no human answers yet.
                                            
                
                    Answered by
            GPT-4o mini
            
    To solve the system of equations:
1. \( 9a + 3b = 30 \) (Equation 1)
2. \( 8a + 4b = 28 \) (Equation 2)
We can simplify both equations.
Starting with Equation 1, we can divide every term by 3:
\[
3a + b = 10 \quad \text{(Equation 3)}
\]
Now, let's simplify Equation 2 by dividing every term by 4:
\[
2a + b = 7 \quad \text{(Equation 4)}
\]
Now we have the simplified system:
1. \( 3a + b = 10 \) (Equation 3)
2. \( 2a + b = 7 \) (Equation 4)
Next, we can eliminate \( b \) by subtracting Equation 4 from Equation 3:
\[
(3a + b) - (2a + b) = 10 - 7
\]
This simplifies to:
\[
3a - 2a + b - b = 3
\]
So, we have:
\[
a = 3
\]
Now, we can substitute this value of \( a \) back into either Equation 3 or Equation 4 to find \( b \). Using Equation 4:
\[
2(3) + b = 7
\]
This becomes:
\[
6 + b = 7
\]
Subtracting 6 from both sides gives:
\[
b = 1
\]
Thus, the solution to the system of equations is:
\[
a = 3, \quad b = 1
\]
    
1. \( 9a + 3b = 30 \) (Equation 1)
2. \( 8a + 4b = 28 \) (Equation 2)
We can simplify both equations.
Starting with Equation 1, we can divide every term by 3:
\[
3a + b = 10 \quad \text{(Equation 3)}
\]
Now, let's simplify Equation 2 by dividing every term by 4:
\[
2a + b = 7 \quad \text{(Equation 4)}
\]
Now we have the simplified system:
1. \( 3a + b = 10 \) (Equation 3)
2. \( 2a + b = 7 \) (Equation 4)
Next, we can eliminate \( b \) by subtracting Equation 4 from Equation 3:
\[
(3a + b) - (2a + b) = 10 - 7
\]
This simplifies to:
\[
3a - 2a + b - b = 3
\]
So, we have:
\[
a = 3
\]
Now, we can substitute this value of \( a \) back into either Equation 3 or Equation 4 to find \( b \). Using Equation 4:
\[
2(3) + b = 7
\]
This becomes:
\[
6 + b = 7
\]
Subtracting 6 from both sides gives:
\[
b = 1
\]
Thus, the solution to the system of equations is:
\[
a = 3, \quad b = 1
\]
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