{:[-11 y=6(z+1)-13 y],[4y-24=c(z-1)]:} For what value of c does the system of linear equations in the variables y and z have infinitely many solutions?

Question

{:[-11 y=6(z+1)-13 y],[4y-24=c(z-1)]:} For what value of  c does the system of linear equations in the variables  y and  z have infinitely many solutions?

Bytelearn · Accepted Answer

Write down equations: Write down the system of linear equations. The system of equations is: -11y = 6(z + 1) - 13y 4y - 24 = c(z - 1)Simplify first equation: Simplify the first equation to express y in terms of z. -11y = 6z + 6 - 13y Combine like terms: -11y + 13y = 6z + 6 2y = 6z + 6 Divide both sides by 2 to solve for y: y = 3z + 3Substitute expression for y: Substitute the expression for y from the first equation into the second equation. 4(3z + 3) - 24 = c(z - 1)Simplify second equation: Simplify the second equation. 12z + 12 - 24 = c(z - 1) 12z - 12 = c(z - 1)Check for proportionality: For the system to have infinitely many solutions, the equations must be proportional. This means that the coefficients of z on both sides of the equation must be equal, and the constants on both sides must also be equal. So, we must have: 12 = c and -12 = -cSolve for c: Solve for c. From the equation 12 = c, we get: c = 12Check constants match: Check if the constants also match when c = 12. -12 = -12 (which is true)Conclude solution: Conclude that the value of c that makes the system have infinitely many solutions is c = 12.

$\begin{aligned} -11 y & =6(z+1)-13 y \\ 4 y-24 & =c(z-1) \end{aligned}$ $\newline$ For what value of $c$ does the system of linear equations in the variables $y$ and $z$ have infinitely many solutions?

Full solution

More problems from Multiply two decimals: where does the decimal point go?

Related topics