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3х + 2y =25
9X + 5y =64​
resolver


Sagot :

Sure, let's solve the system of equations step by step:

Given equations:

1. \( 3x + 2y = 25 \)

2. \( 9x + 5y = 64 \)

We can solve this system using the method of elimination or substitution. Here, I'll demonstrate the substitution method:

**Step 1: Solve for one variable in terms of the other from the first equation.**

From equation (1):

\[ 3x + 2y = 25 \]

Let's solve for \( x \):

\[ 3x = 25 - 2y \]

\[ x = \frac{25 - 2y}{3} \]

**Step 2: Substitute \( x \) from step 1 into the second equation.**

Substitute \( x = \frac{25 - 2y}{3} \) into equation (2):

\[ 9\left(\frac{25 - 2y}{3}\right) + 5y = 64 \]

**Step 3: Simplify and solve for \( y \).**

Multiply through by 3 to eliminate the fraction:

\[ 9(25 - 2y) + 15y = 192 \]

\[ 225 - 18y + 15y = 192 \]

Combine like terms:

\[ 225 - 3y = 192 \]

Subtract 225 from both sides:

\[ -3y = -33 \]

Divide both sides by -3:

\[ y = 11 \]

**Step 4: Substitute \( y = 11 \) back into the equation from step 1 to find \( x \).**

Using \( y = 11 \):

\[ x = \frac{25 - 2 \cdot 11}{3} \]

\[ x = \frac{25 - 22}{3} \]

\[ x = \frac{3}{3} \]

\[ x = 1 \]

**Step 5: Verify the solution by substituting \( x = 1 \) and \( y = 11 \) into both original equations.**

For equation (1):

\[ 3(1) + 2(11) = 3 + 22 = 25 \] (True)

For equation (2):

\[ 9(1) + 5(11) = 9 + 55 = 64 \] (True)

Therefore, the solution to the system of equations is \( x = 1 \) and \( y = 11 \).

**Conclusion:**

{x = 1, \; y = 11}  

Answer: