In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `3x+5y=28`
(ii) `4x+10y=16`

Answer

To solve the given equations step by step, we will follow the method of elimination.

### Step 1: Write down the equations
The two equations are:
1. \( 3x + 5y = 28 \) (Equation I)
2. \( 4x + 10y = 16 \) (Equation II)

### Step 2: Modify Equation I
To eliminate \(y\), we can manipulate the equations. We can multiply Equation I by 2 so that the coefficients of \(y\) in both equations become the same.

\[ 2 \times (3x + 5y) = 2 \times 28 \]
This gives us:
\[ 6x + 10y = 56 \quad \text{(Modified Equation I)} \]

### Step 3: Write down the modified equations
Now we have:
1. \( 6x + 10y = 56 \) (Modified Equation I)
2. \( 4x + 10y = 16 \) (Equation II)

### Step 4: Subtract Equation II from Modified Equation I
Now we will subtract Equation II from the modified Equation I to eliminate \(y\):

\[ (6x + 10y) - (4x + 10y) = 56 - 16 \]
This simplifies to:
\[ 2x = 40 \]

### Step 5: Solve for \(x\)
Now, divide both sides by 2 to solve for \(x\):
\[ x = \frac{40}{2} = 20 \]

### Step 6: Substitute \(x\) back into Equation I
Now that we have the value of \(x\), we can substitute it back into Equation I to find \(y\):
\[ 3(20) + 5y = 28 \]
This simplifies to:
\[ 60 + 5y = 28 \]

### Step 7: Solve for \(y\)
Now, isolate \(y\):
\[ 5y = 28 - 60 \]
\[ 5y = -32 \]
Now, divide by 5:
\[ y = \frac{-32}{5} = -6.4 \]

### Final Solution
Thus, the solution to the equations is:
\[ x = 20, \quad y = -6.4 \]

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