Answer :
Sure, let's solve the system of linear equations step-by-step:
Given the system of equations:
[tex]\[ \begin{array}{r} 3x - 9y = 3 \\ x + 2y = 16 \end{array} \][/tex]
### Step 1: Solve one equation for one variable
We can begin by solving the second equation for [tex]\( x \)[/tex]:
[tex]\[ x + 2y = 16 \][/tex]
Subtract [tex]\( 2y \)[/tex] from both sides:
[tex]\[ x = 16 - 2y \][/tex]
### Step 2: Substitute into the first equation
Substitute [tex]\( x = 16 - 2y \)[/tex] into the first equation:
[tex]\[ 3(16 - 2y) - 9y = 3 \][/tex]
### Step 3: Simplify the equation
Distribute the 3:
[tex]\[ 48 - 6y - 9y = 3 \][/tex]
Combine like terms:
[tex]\[ 48 - 15y = 3 \][/tex]
### Step 4: Solve for [tex]\( y \)[/tex]
Subtract 48 from both sides:
[tex]\[ -15y = 3 - 48 \][/tex]
[tex]\[ -15y = -45 \][/tex]
Divide both sides by -15:
[tex]\[ y = \frac{-45}{-15} \][/tex]
[tex]\[ y = 3 \][/tex]
### Step 5: Substitute back to find [tex]\( x \)[/tex]
Now, substitute [tex]\( y = 3 \)[/tex] back into the equation [tex]\( x = 16 - 2y \)[/tex]:
[tex]\[ x = 16 - 2(3) \][/tex]
[tex]\[ x = 16 - 6 \][/tex]
[tex]\[ x = 10 \][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[ x = 10, \quad y = 3 \][/tex]
Therefore, the values are [tex]\( x = 10 \)[/tex] and [tex]\( y = 3 \)[/tex].
Given the system of equations:
[tex]\[ \begin{array}{r} 3x - 9y = 3 \\ x + 2y = 16 \end{array} \][/tex]
### Step 1: Solve one equation for one variable
We can begin by solving the second equation for [tex]\( x \)[/tex]:
[tex]\[ x + 2y = 16 \][/tex]
Subtract [tex]\( 2y \)[/tex] from both sides:
[tex]\[ x = 16 - 2y \][/tex]
### Step 2: Substitute into the first equation
Substitute [tex]\( x = 16 - 2y \)[/tex] into the first equation:
[tex]\[ 3(16 - 2y) - 9y = 3 \][/tex]
### Step 3: Simplify the equation
Distribute the 3:
[tex]\[ 48 - 6y - 9y = 3 \][/tex]
Combine like terms:
[tex]\[ 48 - 15y = 3 \][/tex]
### Step 4: Solve for [tex]\( y \)[/tex]
Subtract 48 from both sides:
[tex]\[ -15y = 3 - 48 \][/tex]
[tex]\[ -15y = -45 \][/tex]
Divide both sides by -15:
[tex]\[ y = \frac{-45}{-15} \][/tex]
[tex]\[ y = 3 \][/tex]
### Step 5: Substitute back to find [tex]\( x \)[/tex]
Now, substitute [tex]\( y = 3 \)[/tex] back into the equation [tex]\( x = 16 - 2y \)[/tex]:
[tex]\[ x = 16 - 2(3) \][/tex]
[tex]\[ x = 16 - 6 \][/tex]
[tex]\[ x = 10 \][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[ x = 10, \quad y = 3 \][/tex]
Therefore, the values are [tex]\( x = 10 \)[/tex] and [tex]\( y = 3 \)[/tex].