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Sagot :
Sure, let's solve the system of equations step-by-step. We are given the following system of equations:
[tex]\[ \left\{\begin{array}{l} \frac{1}{2} x + \frac{1}{3} y = 5 \\ \frac{1}{4} x + y = 10 \end{array}\right. \][/tex]
First, let's rewrite the equations to make them easier to work with:
[tex]\[ \frac{1}{2} x + \frac{1}{3} y = 5 \quad \text{(1)} \][/tex]
[tex]\[ \frac{1}{4} x + y = 10 \quad \text{(2)} \][/tex]
To eliminate the fractions, we can multiply both equations by the least common multiple (LCM) of the denominators.
For equation (1), the LCM of 2 and 3 is 6:
[tex]\[ 6 \left( \frac{1}{2} x + \frac{1}{3} y \right) = 6 \cdot 5 \][/tex]
[tex]\[ 3x + 2y = 30 \quad \text{(3)} \][/tex]
For equation (2), the LCM of 4 and 1 is 4:
[tex]\[ 4 \left( \frac{1}{4} x + y \right) = 4 \cdot 10 \][/tex]
[tex]\[ x + 4y = 40 \quad \text{(4)} \][/tex]
Now, we have a new system of equations without fractions:
[tex]\[ \left\{\begin{array}{l} 3x + 2y = 30 \\ x + 4y = 40 \end{array}\right. \][/tex]
Next, we can use the method of substitution or elimination. Here, we'll use the elimination method. We want to eliminate one of the variables.
First, let's make the coefficients of [tex]\(x\)[/tex] the same in both equations. We can do this by multiplying equation (4) by 3:
[tex]\[ 3(x + 4y) = 3 \cdot 40 \][/tex]
[tex]\[ 3x + 12y = 120 \quad \text{(5)} \][/tex]
Now we have the system:
[tex]\[ \left\{\begin{array}{l} 3x + 2y = 30 \\ 3x + 12y = 120 \end{array}\right. \][/tex]
Let's subtract equation (3) from equation (5):
[tex]\[ (3x + 12y) - (3x + 2y) = 120 - 30 \][/tex]
[tex]\[ 3x + 12y - 3x - 2y = 90 \][/tex]
[tex]\[ 10y = 90 \][/tex]
[tex]\[ y = \frac{90}{10} \][/tex]
[tex]\[ y = 9 \][/tex]
Now that we have the value of [tex]\(y\)[/tex], we substitute [tex]\(y = 9\)[/tex] back into either of the original equations to solve for [tex]\(x\)[/tex]. We'll use equation (2):
[tex]\[ \frac{1}{4} x + 9 = 10 \][/tex]
[tex]\[ \frac{1}{4} x = 10 - 9 \][/tex]
[tex]\[ \frac{1}{4} x = 1 \][/tex]
[tex]\[ x = 4 \][/tex]
So, the solution to the system of equations is:
[tex]\[ x = 4 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the solution in ordered pair form is:
[tex]\[ \boxed{(4, 9)} \][/tex]
[tex]\[ \left\{\begin{array}{l} \frac{1}{2} x + \frac{1}{3} y = 5 \\ \frac{1}{4} x + y = 10 \end{array}\right. \][/tex]
First, let's rewrite the equations to make them easier to work with:
[tex]\[ \frac{1}{2} x + \frac{1}{3} y = 5 \quad \text{(1)} \][/tex]
[tex]\[ \frac{1}{4} x + y = 10 \quad \text{(2)} \][/tex]
To eliminate the fractions, we can multiply both equations by the least common multiple (LCM) of the denominators.
For equation (1), the LCM of 2 and 3 is 6:
[tex]\[ 6 \left( \frac{1}{2} x + \frac{1}{3} y \right) = 6 \cdot 5 \][/tex]
[tex]\[ 3x + 2y = 30 \quad \text{(3)} \][/tex]
For equation (2), the LCM of 4 and 1 is 4:
[tex]\[ 4 \left( \frac{1}{4} x + y \right) = 4 \cdot 10 \][/tex]
[tex]\[ x + 4y = 40 \quad \text{(4)} \][/tex]
Now, we have a new system of equations without fractions:
[tex]\[ \left\{\begin{array}{l} 3x + 2y = 30 \\ x + 4y = 40 \end{array}\right. \][/tex]
Next, we can use the method of substitution or elimination. Here, we'll use the elimination method. We want to eliminate one of the variables.
First, let's make the coefficients of [tex]\(x\)[/tex] the same in both equations. We can do this by multiplying equation (4) by 3:
[tex]\[ 3(x + 4y) = 3 \cdot 40 \][/tex]
[tex]\[ 3x + 12y = 120 \quad \text{(5)} \][/tex]
Now we have the system:
[tex]\[ \left\{\begin{array}{l} 3x + 2y = 30 \\ 3x + 12y = 120 \end{array}\right. \][/tex]
Let's subtract equation (3) from equation (5):
[tex]\[ (3x + 12y) - (3x + 2y) = 120 - 30 \][/tex]
[tex]\[ 3x + 12y - 3x - 2y = 90 \][/tex]
[tex]\[ 10y = 90 \][/tex]
[tex]\[ y = \frac{90}{10} \][/tex]
[tex]\[ y = 9 \][/tex]
Now that we have the value of [tex]\(y\)[/tex], we substitute [tex]\(y = 9\)[/tex] back into either of the original equations to solve for [tex]\(x\)[/tex]. We'll use equation (2):
[tex]\[ \frac{1}{4} x + 9 = 10 \][/tex]
[tex]\[ \frac{1}{4} x = 10 - 9 \][/tex]
[tex]\[ \frac{1}{4} x = 1 \][/tex]
[tex]\[ x = 4 \][/tex]
So, the solution to the system of equations is:
[tex]\[ x = 4 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the solution in ordered pair form is:
[tex]\[ \boxed{(4, 9)} \][/tex]
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