Let's solve the system of linear equations step by step:
Given the system of equations:
1. [tex]\(4x + 5y = 3\)[/tex]
2. [tex]\(6x - 10y = 1\)[/tex]
### Step 1: Simplify the equations if possible
First, let's rewrite the second equation in standard form:
[tex]\[6x - 10y = 1\][/tex]
### Step 2: Use the Substitution or Elimination Method
We'll use the elimination method for simplicity. We need to eliminate one of the variables by making their coefficients equal in both equations.
Notice that if we multiply the first equation by 3 and the second equation by 2, the coefficients of [tex]\(x\)[/tex] will become equal:
Multiply the first equation by 3:
[tex]\[3(4x + 5y) = 3 \cdot 3\][/tex]
[tex]\[12x + 15y = 9 \tag{3}\][/tex]
Multiply the second equation by 2:
[tex]\[2(6x - 10y) = 2 \cdot 1\][/tex]
[tex]\[12x - 20y = 2 \tag{4}\][/tex]
### Step 3: Subtract one equation from the other
Subtract Equation (4) from Equation (3):
[tex]\[(12x + 15y) - (12x - 20y) = 9 - 2\][/tex]
[tex]\[12x + 15y - 12x + 20y = 7\][/tex]
[tex]\[35y = 7\][/tex]
### Step 4: Solve for [tex]\(y\)[/tex]
Divide both sides by 35 to solve for [tex]\(y\)[/tex]:
[tex]\[y = \frac{7}{35}\][/tex]
[tex]\[y = \frac{1}{5}\][/tex]
### Step 5: Solve for [tex]\(x\)[/tex]
Now that we have [tex]\(y = \frac{1}{5}\)[/tex], substitute [tex]\(y\)[/tex] back into one of the original equations. Let's use the first equation:
[tex]\[4x + 5y = 3\][/tex]
[tex]\[4x + 5\left(\frac{1}{5}\right) = 3\][/tex]
[tex]\[4x + 1 = 3\][/tex]
Subtract 1 from both sides:
[tex]\[4x = 2\][/tex]
Divide both sides by 4:
[tex]\[x = \frac{2}{4}\][/tex]
[tex]\[x = \frac{1}{2}\][/tex]
### Solution:
The solutions for the system of equations are:
[tex]\[x = \frac{1}{2}\][/tex]
[tex]\[y = \frac{1}{5}\][/tex]
Thus, the coordinates [tex]\(\left(\frac{1}{2}, \frac{1}{5}\right)\)[/tex] satisfy both equations in the given system.
