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Sagot :
Certainly! Let's solve the system of linear equations step by step:
[tex]\[ \begin{array}{c} \frac{3x}{2} - \frac{5y}{3} = -2 \quad \text{(1)} \\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad \text{(2)} \end{array} \][/tex]
### Step 1: Clear the denominators
First, let's clear the denominators in both equations to simplify our calculations.
For equation (1):
[tex]\[ \frac{3x}{2} - \frac{5y}{3} = -2 \][/tex]
Multiply every term by the least common multiple (LCM) of the denominators (2 and 3), which is 6:
[tex]\[ 6 \left(\frac{3x}{2}\right) - 6 \left(\frac{5y}{3}\right) = 6(-2) \][/tex]
This simplifies to:
[tex]\[ 9x - 10y = -12 \quad \text{(3)} \][/tex]
For equation (2):
[tex]\[ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \][/tex]
Multiply every term by the LCM of the denominators (3 and 2), which is 6:
[tex]\[ 6 \left(\frac{x}{3}\right) + 6 \left(\frac{y}{2}\right) = 6\left(\frac{13}{6}\right) \][/tex]
This simplifies to:
[tex]\[ 2x + 3y = 13 \quad \text{(4)} \][/tex]
Now, we have the system of equations:
[tex]\[ \begin{array}{c} 9x - 10y = -12 \quad \text{(3)} \\ 2x + 3y = 13 \quad \text{(4)} \end{array} \][/tex]
### Step 2: Solve the system using substitution or elimination
We will use the elimination method to solve this system. First, we will make the coefficients of [tex]\( x \)[/tex] or [tex]\( y \)[/tex] the same in magnitude for both equations.
Let’s eliminate [tex]\( x \)[/tex]. To do this, we need to multiply equation (4) by 9/2 to match the coefficient of [tex]\( x \)[/tex] in equation (3).
Multiply equation (4) by 9/2:
[tex]\[ \left(\frac{9}{2}\right) (2x + 3y) = \left(\frac{9}{2}\right) (13) \][/tex]
This simplifies to:
[tex]\[ 9x + \frac{27y}{2} = 58.5 \quad \text{(5)} \][/tex]
Now, rewrite equation (3) for convenience:
[tex]\[ 9x - 10y = -12 \quad \text{(3)} \][/tex]
### Step 3: Eliminate and solve for [tex]\( y \)[/tex]
Subtract equation (3) from equation (5):
[tex]\[ (9x + \frac{27y}{2}) - (9x - 10y) = 58.5 - (-12) \][/tex]
Simplifying, we have:
[tex]\[ 9x + \frac{27y}{2} - 9x + 10y = 58.5 + 12 \][/tex]
This reduces to:
[tex]\[ \frac{27y}{2} + 10y = 70.5 \][/tex]
Combine the terms containing [tex]\( y \)[/tex]:
[tex]\[ \frac{27y}{2} + 10y = \frac{70y + 27y}{2} = 70.5 \][/tex]
[tex]\[ \frac{47y}{2} = 70.5 \][/tex]
Multiply both sides by 2:
[tex]\[ 47y = 141 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{141}{47} = 3 \][/tex]
So, [tex]\( y = 3 \)[/tex].
### Step 4: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]
Substitute [tex]\( y = 3 \)[/tex] back into equation (4):
[tex]\[ 2x + 3(3) = 13 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 9 = 13 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, [tex]\( x = 2 \)[/tex].
### Conclusion:
The solution to the system of equations is:
[tex]\[ x = 2 \quad \text{and} \quad y = 3 \][/tex]
Thus, the solution is [tex]\((x, y) = (2, 3)\)[/tex].
[tex]\[ \begin{array}{c} \frac{3x}{2} - \frac{5y}{3} = -2 \quad \text{(1)} \\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad \text{(2)} \end{array} \][/tex]
### Step 1: Clear the denominators
First, let's clear the denominators in both equations to simplify our calculations.
For equation (1):
[tex]\[ \frac{3x}{2} - \frac{5y}{3} = -2 \][/tex]
Multiply every term by the least common multiple (LCM) of the denominators (2 and 3), which is 6:
[tex]\[ 6 \left(\frac{3x}{2}\right) - 6 \left(\frac{5y}{3}\right) = 6(-2) \][/tex]
This simplifies to:
[tex]\[ 9x - 10y = -12 \quad \text{(3)} \][/tex]
For equation (2):
[tex]\[ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \][/tex]
Multiply every term by the LCM of the denominators (3 and 2), which is 6:
[tex]\[ 6 \left(\frac{x}{3}\right) + 6 \left(\frac{y}{2}\right) = 6\left(\frac{13}{6}\right) \][/tex]
This simplifies to:
[tex]\[ 2x + 3y = 13 \quad \text{(4)} \][/tex]
Now, we have the system of equations:
[tex]\[ \begin{array}{c} 9x - 10y = -12 \quad \text{(3)} \\ 2x + 3y = 13 \quad \text{(4)} \end{array} \][/tex]
### Step 2: Solve the system using substitution or elimination
We will use the elimination method to solve this system. First, we will make the coefficients of [tex]\( x \)[/tex] or [tex]\( y \)[/tex] the same in magnitude for both equations.
Let’s eliminate [tex]\( x \)[/tex]. To do this, we need to multiply equation (4) by 9/2 to match the coefficient of [tex]\( x \)[/tex] in equation (3).
Multiply equation (4) by 9/2:
[tex]\[ \left(\frac{9}{2}\right) (2x + 3y) = \left(\frac{9}{2}\right) (13) \][/tex]
This simplifies to:
[tex]\[ 9x + \frac{27y}{2} = 58.5 \quad \text{(5)} \][/tex]
Now, rewrite equation (3) for convenience:
[tex]\[ 9x - 10y = -12 \quad \text{(3)} \][/tex]
### Step 3: Eliminate and solve for [tex]\( y \)[/tex]
Subtract equation (3) from equation (5):
[tex]\[ (9x + \frac{27y}{2}) - (9x - 10y) = 58.5 - (-12) \][/tex]
Simplifying, we have:
[tex]\[ 9x + \frac{27y}{2} - 9x + 10y = 58.5 + 12 \][/tex]
This reduces to:
[tex]\[ \frac{27y}{2} + 10y = 70.5 \][/tex]
Combine the terms containing [tex]\( y \)[/tex]:
[tex]\[ \frac{27y}{2} + 10y = \frac{70y + 27y}{2} = 70.5 \][/tex]
[tex]\[ \frac{47y}{2} = 70.5 \][/tex]
Multiply both sides by 2:
[tex]\[ 47y = 141 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{141}{47} = 3 \][/tex]
So, [tex]\( y = 3 \)[/tex].
### Step 4: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]
Substitute [tex]\( y = 3 \)[/tex] back into equation (4):
[tex]\[ 2x + 3(3) = 13 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 9 = 13 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, [tex]\( x = 2 \)[/tex].
### Conclusion:
The solution to the system of equations is:
[tex]\[ x = 2 \quad \text{and} \quad y = 3 \][/tex]
Thus, the solution is [tex]\((x, y) = (2, 3)\)[/tex].
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