Answer :
Let's solve the given system of linear equations step by step:
[tex]\[ \left\{ \begin{array}{c} \frac{2}{3} x - \frac{3}{4} y = \frac{1}{6} \\ \frac{1}{8} x - \frac{5}{6} y = 12 \end{array} \right. \][/tex]
Let's label them as Equation 1 and Equation 2:
Equation 1:
[tex]\[ \frac{2}{3} x - \frac{3}{4} y = \frac{1}{6} \][/tex]
Equation 2:
[tex]\[ \frac{1}{8} x - \frac{5}{6} y = 12 \][/tex]
Step 1: Clear fractions by finding the common denominators.
For Equation 1, the common denominator of 3, 4, and 6 is 12, so we multiply every term by 12:
[tex]\[ 12 \left(\frac{2}{3} x\right) - 12 \left(\frac{3}{4} y\right) = 12 \left(\frac{1}{6}\right) \][/tex]
This simplifies to:
[tex]\[ 8x - 9y = 2 \][/tex]
For Equation 2, the common denominator of 8 and 6 is 24, so we multiply every term by 24:
[tex]\[ 24 \left(\frac{1}{8} x\right) - 24 \left(\frac{5}{6} y\right) = 24 \cdot 12 \][/tex]
This simplifies to:
[tex]\[ 3x - 20y = 288 \][/tex]
Now we have the system:
[tex]\[ \left\{ \begin{array}{c} 8x - 9y = 2 \\ 3x - 20y = 288 \end{array} \right. \][/tex]
Step 2: Use the method of substitution or elimination to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Here we'll use the elimination method.
First, we manipulate the equations for easier elimination. Multiply Equation 1 by 3 and Equation 2 by 8:
[tex]\[ 3(8x - 9y) = 3 \cdot 2 \][/tex]
[tex]\[ 8(3x - 20y) = 8 \cdot 288 \][/tex]
This gives us:
[tex]\[ 24x - 27y = 6 \][/tex]
[tex]\[ 24x - 160y = 2304 \][/tex]
Now subtract the first modified equation from the second one:
[tex]\[ (24x - 160y) - (24x - 27y) = 2304 - 6 \][/tex]
This reduces to:
[tex]\[ -133y = 2298 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{2298}{-133} \approx -17.278 \][/tex]
Step 3: Substitute [tex]\(y\)[/tex] back into one of the original simplified equations to find [tex]\(x\)[/tex]. We use the first simplified equation:
[tex]\[ 8x - 9(-17.278) = 2 \][/tex]
This simplifies to:
[tex]\[ 8x + 155.502 = 2 \][/tex]
Subtract 155.502 from both sides:
[tex]\[ 8x = 2 - 155.502 \][/tex]
[tex]\[ 8x = -153.502 \][/tex]
Dividing by 8:
[tex]\[ x = \frac{-153.502}{8} \approx -19.188 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x \approx -19.188 \][/tex]
[tex]\[ y \approx -17.278 \][/tex]
[tex]\[ \left\{ \begin{array}{c} \frac{2}{3} x - \frac{3}{4} y = \frac{1}{6} \\ \frac{1}{8} x - \frac{5}{6} y = 12 \end{array} \right. \][/tex]
Let's label them as Equation 1 and Equation 2:
Equation 1:
[tex]\[ \frac{2}{3} x - \frac{3}{4} y = \frac{1}{6} \][/tex]
Equation 2:
[tex]\[ \frac{1}{8} x - \frac{5}{6} y = 12 \][/tex]
Step 1: Clear fractions by finding the common denominators.
For Equation 1, the common denominator of 3, 4, and 6 is 12, so we multiply every term by 12:
[tex]\[ 12 \left(\frac{2}{3} x\right) - 12 \left(\frac{3}{4} y\right) = 12 \left(\frac{1}{6}\right) \][/tex]
This simplifies to:
[tex]\[ 8x - 9y = 2 \][/tex]
For Equation 2, the common denominator of 8 and 6 is 24, so we multiply every term by 24:
[tex]\[ 24 \left(\frac{1}{8} x\right) - 24 \left(\frac{5}{6} y\right) = 24 \cdot 12 \][/tex]
This simplifies to:
[tex]\[ 3x - 20y = 288 \][/tex]
Now we have the system:
[tex]\[ \left\{ \begin{array}{c} 8x - 9y = 2 \\ 3x - 20y = 288 \end{array} \right. \][/tex]
Step 2: Use the method of substitution or elimination to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Here we'll use the elimination method.
First, we manipulate the equations for easier elimination. Multiply Equation 1 by 3 and Equation 2 by 8:
[tex]\[ 3(8x - 9y) = 3 \cdot 2 \][/tex]
[tex]\[ 8(3x - 20y) = 8 \cdot 288 \][/tex]
This gives us:
[tex]\[ 24x - 27y = 6 \][/tex]
[tex]\[ 24x - 160y = 2304 \][/tex]
Now subtract the first modified equation from the second one:
[tex]\[ (24x - 160y) - (24x - 27y) = 2304 - 6 \][/tex]
This reduces to:
[tex]\[ -133y = 2298 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{2298}{-133} \approx -17.278 \][/tex]
Step 3: Substitute [tex]\(y\)[/tex] back into one of the original simplified equations to find [tex]\(x\)[/tex]. We use the first simplified equation:
[tex]\[ 8x - 9(-17.278) = 2 \][/tex]
This simplifies to:
[tex]\[ 8x + 155.502 = 2 \][/tex]
Subtract 155.502 from both sides:
[tex]\[ 8x = 2 - 155.502 \][/tex]
[tex]\[ 8x = -153.502 \][/tex]
Dividing by 8:
[tex]\[ x = \frac{-153.502}{8} \approx -19.188 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x \approx -19.188 \][/tex]
[tex]\[ y \approx -17.278 \][/tex]
