To solve the system of equations by the substitution method, we will follow these steps:
Given Equations:
1. [tex]\( y = \frac{4}{7}x + \frac{3}{7} \)[/tex]
2. [tex]\( y = \frac{3}{4}x - 1 \)[/tex]
Step 1: Substituting Equation 1 into Equation 2.
Since both equations are equal to [tex]\( y \)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[ \frac{4}{7}x + \frac{3}{7} = \frac{3}{4}x - 1 \][/tex]
Step 2: Eliminating the Fractions.
To eliminate the fractions, find the least common multiple (LCM) of the denominators, which are 7 and 4. The LCM is 28. Multiply every term in the equation by 28:
[tex]\[ 28 \left( \frac{4}{7}x + \frac{3}{7} \right) = 28 \left( \frac{3}{4}x - 1 \right) \][/tex]
This results in:
[tex]\[ 4x \cdot 4 + 4 \cdot 3 = 7 \cdot 3x - 28 \][/tex]
which simplifies to:
[tex]\[ 16x + 12 = 21x - 28 \][/tex]
Step 3: Solving for [tex]\( x \)[/tex].
Now, let's solve for [tex]\( x \)[/tex]:
First, get all [tex]\( x \)[/tex]-terms on one side and constants on the other:
[tex]\[ 16x + 12 = 21x - 28 \][/tex]
Subtract [tex]\( 16x \)[/tex] from both sides:
[tex]\[ 12 = 5x - 28 \][/tex]
Next, add 28 to both sides:
[tex]\[ 12 + 28 = 5x \][/tex]
[tex]\[ 40 = 5x \][/tex]
Finally, divide by 5:
[tex]\[ x = \frac{40}{5} = 8 \][/tex]
So, [tex]\( x = 8 \)[/tex].
Step 4: Substituting [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex].
Now substitute [tex]\( x = 8 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Using the first equation:
[tex]\[ y = \frac{4}{7}x + \frac{3}{7} \][/tex]
[tex]\[ y = \frac{4}{7}(8) + \frac{3}{7} \][/tex]
First, calculate [tex]\( \frac{4}{7} \times 8 \)[/tex]:
[tex]\[ \frac{4 \times 8}{7} = \frac{32}{7} \][/tex]
Then, add [tex]\( \frac{3}{7} \)[/tex]:
[tex]\[ y = \frac{32}{7} + \frac{3}{7} = \frac{35}{7} = 5 \][/tex]
So, [tex]\( y = 5 \)[/tex].
Solution:
The solution to the system of equations is [tex]\( \mathbf{(x, y) = (8, 5)} \)[/tex].
