Question

d 22x + 40y = -14 et 44x - 4y = 8
(44x*(22x + 40y = -14))
(22x*(-44x - 4y = 8))
968x + 1760 = 616
-968 - 88 = -1056
4848 = 492y/492
réponse y = 2,33

Explanation:

Step1: Correct the system of equations

The original equations seem to have a sign error. Let's assume the correct system is \(

$$\begin{cases} 22x + 40y = - 14 \\ 44x - 4y = 8 \end{cases}$$

\) (correcting the signs as the original handwritten might have typos). To eliminate \(x\), multiply the first equation by \(2\): \(2\times(22x + 40y)=2\times(-14)\), which gives \(44x + 80y=-28\).

Step2: Subtract the second equation

Now we have \(44x + 80y=-28\) and \(44x - 4y = 8\). Subtract the second equation from the first: \((44x + 80y)-(44x - 4y)=-28 - 8\). Simplify the left - hand side: \(44x+80y - 44x + 4y=-36\), so \(84y=-36\).

Step3: Solve for \(y\)

Divide both sides of \(84y=-36\) by \(84\): \(y=\frac{-36}{84}=\frac{-3}{7}\approx - 0.4286\). Wait, the handwritten answer has \(y = 2.33\) which is incorrect. Let's check the original equations again. Maybe the first equation is \(22x+40y = 14\) (without the negative sign). Then multiply the first equation by \(2\): \(44x + 80y = 28\). Subtract the second equation \(44x-4y = 8\) from it: \((44x + 80y)-(44x - 4y)=28 - 8\), \(84y = 20\), \(y=\frac{20}{84}=\frac{5}{21}\approx0.238\). Still not \(2.33\). Maybe the first equation is \(22x + 40y=-14\) and the second is \(- 44x-4y = 8\) (multiplying the second equation by \(- 1\)). Then add the two equations: \((22x + 40y)+(-44x-4y)=-14 + 8\), \(-22x+36y=-6\). This is getting more complicated. Let's use the correct elimination method for the system \(

$$\begin{cases}22x + 40y=14\\44x-4y = 8\end{cases}$$

\) (assuming the first equation is positive). Multiply the first equation by \(2\): \(44x + 80y = 28\). Subtract the second equation: \((44x + 80y)-(44x - 4y)=28 - 8\), \(84y = 20\), \(y=\frac{20}{84}=\frac{5}{21}\approx0.238\). If we take the system as \(

$$\begin{cases}22x+40y=-14\\-44x - 4y=8\end{cases}$$

\) (multiplying the second equation by \(-1\) to make \(x\) coefficients opposite in sign for addition). Add the two equations: \(22x+40y-44x - 4y=-14 + 8\), \(-22x + 36y=-6\). Let's solve for \(x\) from the second original equation \(44x-4y = 8\), \(44x=8 + 4y\), \(x=\frac{8 + 4y}{44}=\frac{2 + y}{11}\). Substitute into the first equation \(22\times\frac{2 + y}{11}+40y=-14\), \(2(2 + y)+40y=-14\), \(4 + 2y+40y=-14\), \(42y=-18\), \(y=\frac{-18}{42}=\frac{-3}{7}\approx - 0.4286\). The handwritten solution has calculation errors. Let's do it properly.

Let's start over with the system \(

$$\begin{cases}22x + 40y=14\ \ (1)\\44x-4y = 8\ \ (2)\end{cases}$$

\)

Step1: Eliminate \(x\)

Multiply equation \((1)\) by \(2\): \(44x+80y = 28\ \ (3)\)

Step2: Subtract equation \((2)\) from \((3)\)

\((44x + 80y)-(44x - 4y)=28 - 8\)

\(44x+80y - 44x + 4y=20\)

\(84y = 20\)

Step3: Solve for \(y\)

\(y=\frac{20}{84}=\frac{5}{21}\approx0.238\)

If the system is \(

$$\begin{cases}22x+40y=-14\ \ (1)\\44x - 4y=8\ \ (2)\end{cases}$$

\)

Step1: Eliminate \(x\)

Multiply equation \((1)\) by \(2\): \(44x + 80y=-28\ \ (3)\)

Step2: Subtract equation \((2)\) from \((3)\)

\((44x + 80y)-(44x - 4y)=-28 - 8\)

\(44x+80y - 44x + 4y=-36\)

\(84y=-36\)

Step3: Solve for \(y\)

\(y=\frac{-36}{84}=\frac{-3}{7}\approx - 0.4286\)

The handwritten answer \(y = 2.33\) is incorrect. The correct value of \(y\) depends on the correct system of equations, but following the proper elimination method, we can see the handwritten solution has arithmetic errors.

Answer:

The handwritten solution has errors. The correct value of \(y\) (depending on the correct system) is either \(\frac{5}{21}\approx0.238\) or \(\frac{-3}{7}\approx - 0.4286\), not \(2.33\).