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x=-(3-cy)+6-7x

3x-y=(3)/(2)-x
In the system of equations, 
c is a constant. For what value of 
c does the system of linear equations have infinitely many solutions?

x=(3cy)+67x x=-(3-c y)+6-7 x \newline3xy=32x 3 x-y=\frac{3}{2}-x \newlineIn the system of equations, c c is a constant. For what value of c c does the system of linear equations have infinitely many solutions?

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Q. x=(3cy)+67x x=-(3-c y)+6-7 x \newline3xy=32x 3 x-y=\frac{3}{2}-x \newlineIn the system of equations, c c is a constant. For what value of c c does the system of linear equations have infinitely many solutions?
  1. Simplify first equation: Let's first simplify the first equation:\newlinex=(3cy)+67xx = -(3 - cy) + 6 - 7x\newlineCombine like terms and move all xx terms to one side:\newlinex+7x=(3+cy)+6x + 7x = -(-3 + cy) + 6\newline8x=3cy+68x = 3 - cy + 6\newline8x=9cy8x = 9 - cy\newlineNow, divide by 88 to solve for xx:\newlinex=9cy8x = \frac{9 - cy}{8}
  2. Simplify second equation: Now let's simplify the second equation:\newline3xy=32x3x - y = \frac{3}{2} - x\newlineMove the x term to the left side to combine with the 3x3x term:\newline3x+x=32+y3x + x = \frac{3}{2} + y\newline4x=32+y4x = \frac{3}{2} + y\newlineNow, divide by 44 to solve for xx:\newlinex=(32+y)4x = \frac{(\frac{3}{2} + y)}{4}\newlinex=38+y4x = \frac{3}{8} + \frac{y}{4}
  3. Set equations equal to each other: For the system to have infinitely many solutions, the two expressions we found for xx must be identical. Therefore, we set them equal to each other:\newline(9cy8)=38+y4(\frac{9 - cy}{8}) = \frac{3}{8} + \frac{y}{4}\newlineTo solve for cc, we need to get rid of the denominators by finding a common denominator and multiplying both sides by it. The common denominator for 88 and 44 is 88, so we multiply both sides by 88:\newline8×(9cy8)=8×(38+y4)8 \times (\frac{9 - cy}{8}) = 8 \times (\frac{3}{8} + \frac{y}{4})\newlineThis simplifies to:\newline9cy=3+2y9 - cy = 3 + 2y
  4. Solve for c: Now, we need to solve for c. First, let's move all terms involving yy to one side and constants to the other: cy2y=39-cy - 2y = 3 - 9 Combine like terms: y(c2)=6y(-c - 2) = -6 Now, divide by yy to solve for cc: c2=6y-c - 2 = -\frac{6}{y} c=6y+2-c = -\frac{6}{y} + 2 Multiply by 1-1 to solve for cc: c=6y2c = \frac{6}{y} - 2
  5. Reconsidering the mistake: However, we made a mistake in our reasoning. For the system to have infinitely many solutions, the equations must be dependent, which means they must be multiples of each other. The mistake was in assuming we could solve for a specific value of cc without considering the relationship between the coefficients of the two equations. We need to go back and compare the coefficients of xx and yy in both equations to find the condition for cc.

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