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Sagot :
To solve the given system of linear equations:
[tex]\[ \left\{ \begin{array}{l} 7x = 5y + 45 \\ 2x - 5y = 4 \end{array} \right. \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously. Here's a step-by-step solution:
1. Isolate one variable in one of the equations.
Let's isolate [tex]\(x\)[/tex] in the first equation:
[tex]\[ 7x = 5y + 45 \][/tex]
Divide both sides by 7 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{5y + 45}{7} \][/tex]
2. Substitute the expression for [tex]\(x\)[/tex] into the second equation.
Substitute [tex]\(x = \frac{5y + 45}{7}\)[/tex] into the second equation [tex]\(2x - 5y = 4\)[/tex]:
[tex]\[ 2\left(\frac{5y + 45}{7}\right) - 5y = 4 \][/tex]
Multiply both sides of the equation by 7 to clear the fraction:
[tex]\[ 2(5y + 45) - 35y = 28 \][/tex]
3. Simplify and solve for [tex]\(y\)[/tex].
Distribute the 2 on the left-hand side:
[tex]\[ 10y + 90 - 35y = 28 \][/tex]
Combine like terms:
[tex]\[ 10y - 35y + 90 = 28 \implies -25y + 90 = 28 \][/tex]
Subtract 90 from both sides:
[tex]\[ -25y = 28 - 90 \implies -25y = -62 \][/tex]
Divide both sides by -25 to solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{62}{25} \][/tex]
4. Substitute the value of [tex]\(y\)[/tex] back into the expression for [tex]\(x\)[/tex].
We already have [tex]\(x = \frac{5y + 45}{7}\)[/tex]. Substitute [tex]\(y = \frac{62}{25}\)[/tex] into this expression:
[tex]\[ x = \frac{5 \left( \frac{62}{25} \right) + 45}{7} \][/tex]
Simplify the numerator:
[tex]\[ x = \frac{\frac{310}{25} + 45}{7} = \frac{\frac{310}{25} + \frac{1125}{25}}{7} = \frac{\frac{1435}{25}}{7} \][/tex]
Combine terms in the numerator:
[tex]\[ x = \frac{1435}{25 \cdot 7} = \frac{1435}{175} = \frac{41}{5} \][/tex]
5. Conclude with the solution.
Therefore, the solution to the system of equations is:
[tex]\[ x = \frac{41}{5}, \quad y = \frac{62}{25} \][/tex]
These values satisfy both equations in the system, providing the solution:
[tex]\[ \left( x, y \right) = \left( \frac{41}{5}, \frac{62}{25} \right) \][/tex]
[tex]\[ \left\{ \begin{array}{l} 7x = 5y + 45 \\ 2x - 5y = 4 \end{array} \right. \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously. Here's a step-by-step solution:
1. Isolate one variable in one of the equations.
Let's isolate [tex]\(x\)[/tex] in the first equation:
[tex]\[ 7x = 5y + 45 \][/tex]
Divide both sides by 7 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{5y + 45}{7} \][/tex]
2. Substitute the expression for [tex]\(x\)[/tex] into the second equation.
Substitute [tex]\(x = \frac{5y + 45}{7}\)[/tex] into the second equation [tex]\(2x - 5y = 4\)[/tex]:
[tex]\[ 2\left(\frac{5y + 45}{7}\right) - 5y = 4 \][/tex]
Multiply both sides of the equation by 7 to clear the fraction:
[tex]\[ 2(5y + 45) - 35y = 28 \][/tex]
3. Simplify and solve for [tex]\(y\)[/tex].
Distribute the 2 on the left-hand side:
[tex]\[ 10y + 90 - 35y = 28 \][/tex]
Combine like terms:
[tex]\[ 10y - 35y + 90 = 28 \implies -25y + 90 = 28 \][/tex]
Subtract 90 from both sides:
[tex]\[ -25y = 28 - 90 \implies -25y = -62 \][/tex]
Divide both sides by -25 to solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{62}{25} \][/tex]
4. Substitute the value of [tex]\(y\)[/tex] back into the expression for [tex]\(x\)[/tex].
We already have [tex]\(x = \frac{5y + 45}{7}\)[/tex]. Substitute [tex]\(y = \frac{62}{25}\)[/tex] into this expression:
[tex]\[ x = \frac{5 \left( \frac{62}{25} \right) + 45}{7} \][/tex]
Simplify the numerator:
[tex]\[ x = \frac{\frac{310}{25} + 45}{7} = \frac{\frac{310}{25} + \frac{1125}{25}}{7} = \frac{\frac{1435}{25}}{7} \][/tex]
Combine terms in the numerator:
[tex]\[ x = \frac{1435}{25 \cdot 7} = \frac{1435}{175} = \frac{41}{5} \][/tex]
5. Conclude with the solution.
Therefore, the solution to the system of equations is:
[tex]\[ x = \frac{41}{5}, \quad y = \frac{62}{25} \][/tex]
These values satisfy both equations in the system, providing the solution:
[tex]\[ \left( x, y \right) = \left( \frac{41}{5}, \frac{62}{25} \right) \][/tex]
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