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Sagot :
To solve the system of equations:
[tex]\[ \left\{ \begin{array}{c} 3y = 4x + 7 \\ -4x - 4y = 28 \end{array} \right. \][/tex]
we will substitute solutions into the equations to verify them.
## Step-by-Step Solution:
### Step 1: Write down the equations in the system:
[tex]\[ \begin{array}{l} 3y = 4x + 7 \quad \text{(Equation 1)} \\ -4x - 4y = 28 \quad \text{(Equation 2)} \end{array} \][/tex]
### Step 2: Solve one of the equations for one variable:
Let's isolate [tex]\( y \)[/tex] in Equation 1:
[tex]\[ 3y = 4x + 7 \][/tex]
[tex]\[ y = \frac{4x + 7}{3} \][/tex]
### Step 3: Substitute the expression for [tex]\( y \)[/tex] into Equation 2:
Substitute [tex]\( y = \frac{4x + 7}{3} \)[/tex] into Equation 2:
[tex]\[ -4x - 4 \left(\frac{4x + 7}{3}\right) = 28 \][/tex]
### Step 4: Clear the fraction by multiplying through by 3:
[tex]\[ -12x - 4(4x + 7) = 84 \][/tex]
[tex]\[ -12x - 16x - 28 = 84 \][/tex]
### Step 5: Combine like terms:
[tex]\[ -28x - 28 = 84 \][/tex]
### Step 6: Solve for [tex]\( x \)[/tex]:
First, add 28 to both sides:
[tex]\[ -28x - 28 + 28 = 84 + 28 \][/tex]
[tex]\[ -28x = 112 \][/tex]
Next, divide by [tex]\(-28\)[/tex]:
[tex]\[ x = \frac{112}{-28} \][/tex]
[tex]\[ x = -4 \][/tex]
### Step 7: Substitute [tex]\( x = -4 \)[/tex] back into the expression we found for [tex]\( y \)[/tex]:
Recall [tex]\( y = \frac{4x + 7}{3} \)[/tex]:
[tex]\[ y = \frac{4(-4) + 7}{3} \][/tex]
[tex]\[ y = \frac{-16 + 7}{3} \][/tex]
[tex]\[ y = \frac{-9}{3} \][/tex]
[tex]\[ y = -3 \][/tex]
### Step 8: Write down the solution:
The solution to the system of equations is:
[tex]\[ x = -4, \quad y = -3 \][/tex]
### Step 9: Verify the solution:
Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -3 \)[/tex] back into the original equations to check.
For [tex]\( 3y = 4x + 7 \)[/tex]:
[tex]\[ 3(-3) = 4(-4) + 7 \][/tex]
[tex]\[ -9 = -16 + 7 \][/tex]
[tex]\[ -9 = -9 \quad \text{(True)} \][/tex]
For [tex]\( -4x - 4y = 28 \)[/tex]:
[tex]\[ -4(-4) - 4(-3) = 28 \][/tex]
[tex]\[ 16 + 12 = 28 \][/tex]
[tex]\[ 28 = 28 \quad \text{(True)} \][/tex]
Since both equations are satisfied, the solution is correct. Therefore, the solution to the system of equations is:
[tex]\[ \boxed{x = -4, y = -3} \][/tex]
[tex]\[ \left\{ \begin{array}{c} 3y = 4x + 7 \\ -4x - 4y = 28 \end{array} \right. \][/tex]
we will substitute solutions into the equations to verify them.
## Step-by-Step Solution:
### Step 1: Write down the equations in the system:
[tex]\[ \begin{array}{l} 3y = 4x + 7 \quad \text{(Equation 1)} \\ -4x - 4y = 28 \quad \text{(Equation 2)} \end{array} \][/tex]
### Step 2: Solve one of the equations for one variable:
Let's isolate [tex]\( y \)[/tex] in Equation 1:
[tex]\[ 3y = 4x + 7 \][/tex]
[tex]\[ y = \frac{4x + 7}{3} \][/tex]
### Step 3: Substitute the expression for [tex]\( y \)[/tex] into Equation 2:
Substitute [tex]\( y = \frac{4x + 7}{3} \)[/tex] into Equation 2:
[tex]\[ -4x - 4 \left(\frac{4x + 7}{3}\right) = 28 \][/tex]
### Step 4: Clear the fraction by multiplying through by 3:
[tex]\[ -12x - 4(4x + 7) = 84 \][/tex]
[tex]\[ -12x - 16x - 28 = 84 \][/tex]
### Step 5: Combine like terms:
[tex]\[ -28x - 28 = 84 \][/tex]
### Step 6: Solve for [tex]\( x \)[/tex]:
First, add 28 to both sides:
[tex]\[ -28x - 28 + 28 = 84 + 28 \][/tex]
[tex]\[ -28x = 112 \][/tex]
Next, divide by [tex]\(-28\)[/tex]:
[tex]\[ x = \frac{112}{-28} \][/tex]
[tex]\[ x = -4 \][/tex]
### Step 7: Substitute [tex]\( x = -4 \)[/tex] back into the expression we found for [tex]\( y \)[/tex]:
Recall [tex]\( y = \frac{4x + 7}{3} \)[/tex]:
[tex]\[ y = \frac{4(-4) + 7}{3} \][/tex]
[tex]\[ y = \frac{-16 + 7}{3} \][/tex]
[tex]\[ y = \frac{-9}{3} \][/tex]
[tex]\[ y = -3 \][/tex]
### Step 8: Write down the solution:
The solution to the system of equations is:
[tex]\[ x = -4, \quad y = -3 \][/tex]
### Step 9: Verify the solution:
Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -3 \)[/tex] back into the original equations to check.
For [tex]\( 3y = 4x + 7 \)[/tex]:
[tex]\[ 3(-3) = 4(-4) + 7 \][/tex]
[tex]\[ -9 = -16 + 7 \][/tex]
[tex]\[ -9 = -9 \quad \text{(True)} \][/tex]
For [tex]\( -4x - 4y = 28 \)[/tex]:
[tex]\[ -4(-4) - 4(-3) = 28 \][/tex]
[tex]\[ 16 + 12 = 28 \][/tex]
[tex]\[ 28 = 28 \quad \text{(True)} \][/tex]
Since both equations are satisfied, the solution is correct. Therefore, the solution to the system of equations is:
[tex]\[ \boxed{x = -4, y = -3} \][/tex]
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