Question7. $6 x-5 y=12$
8. $y=6 x-8$

Studdy Solution

STEP 1

1. We are given a system of two linear equations: - Equation 1: $6x - 5y = 12$ - Equation 2: $y = 6x - 8$
2. We need to find the values of $x$ and $y$ that satisfy both equations simultaneously.
3. We will use the substitution method to solve this system of equations.

STEP 2

1. Substitute Equation 2 into Equation 1.
2. Solve for $x$ .
3. Substitute the value of $x$ back into Equation 2 to find $y$ .
4. Verify the solution by checking both equations.

STEP 3

Substitute the expression for $y$ from Equation 2 into Equation 1:
$6x - 5(6x - 8) = 12$

STEP 4

Distribute the $-5$ in the equation:
$6x - 30x + 40 = 12$
Combine like terms:
$-24x + 40 = 12$

STEP 5

Subtract 40 from both sides to isolate the term with $x$ :
$-24x = 12 - 40$
$-24x = -28$
Divide both sides by $-24$ to solve for $x$ :
$x = \frac{-28}{-24}$
Simplify the fraction:
$x = \frac{7}{6}$

STEP 6

Substitute $x = \frac{7}{6}$ back into Equation 2 to solve for $y$ :
$y = 6\left(\frac{7}{6}\right) - 8$
Simplify:
$y = 7 - 8$
$y = -1$

STEP 7

Verify the solution by substituting $x = \frac{7}{6}$ and $y = -1$ into both original equations.
For Equation 1:
$6\left(\frac{7}{6}\right) - 5(-1) = 12$
Simplify:
$7 + 5 = 12$
$12 = 12$ (True)
For Equation 2:
$y = 6\left(\frac{7}{6}\right) - 8$
$-1 = 7 - 8$
$-1 = -1$ (True)
Both equations are satisfied.
The solution is:
$x = \frac{7}{6}, \quad y = -1$

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