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1. effectuez les multiplications nécessaires, celles mentionnées à léta…

Question

  1. effectuez les multiplications nécessaires, celles mentionnées à létape 2 de lencadré théorique précédent, pour obtenir deux équations permettant dutiliser la méthode de réduction.

a ( 2x + 4y = -1 ) et ( 6x - y = 10 )
d ( 22x + 40y = 14 ) et ( 44x - 4y = 8 )

Explanation:

Step1: Analyze System (a)

We have the system \(

$$\begin{cases} 2x + 4y=- 1\\6x - y = 10\end{cases}$$

\). To use elimination, we can multiply the second equation by 4 to make the coefficients of \(y\) opposite. Multiply \(6x - y=10\) by 4: \(4\times(6x - y)=4\times10\) gives \(24x-4y = 40\). Now the system is \(

$$\begin{cases}2x + 4y=-1\\24x-4y = 40\end{cases}$$

\).

Step2: Analyze System (d)

We have the system \(

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

\). To use elimination, we can multiply the second equation by 10 to make the coefficients of \(y\) opposite. Multiply \(44x - 4y=8\) by 10: \(10\times(44x-4y)=10\times8\) gives \(440x-40y = 80\). Now the system is \(

$$\begin{cases}22x + 40y=14\\440x-40y = 80\end{cases}$$

\).

(For part (a), after multiplying the second equation by 4, we add the two equations: \((2x + 4y)+(24x-4y)=-1 + 40\), \(26x=39\), \(x=\frac{39}{26}=\frac{3}{2}\). Substitute \(x = \frac{3}{2}\) into \(6x-y = 10\): \(6\times\frac{3}{2}-y=10\), \(9 - y=10\), \(y=-1\). So the solution is \((\frac{3}{2},-1)\). For part (d), after multiplying the second equation by 10, we add the two equations: \((22x + 40y)+(440x-40y)=14 + 80\), \(462x=94\), \(x=\frac{94}{462}=\frac{47}{231}\approx0.203\). Substitute \(x\) into \(44x-4y = 8\) to find \(y\), but the hand - written answer for \(y\) seems incorrect. However, the main task was to perform the multiplications for elimination, which we did for both systems.)

Answer:

For system (a): Multiply \(6x - y = 10\) by 4 to get \(24x-4y = 40\), resulting in the system \(

$$\begin{cases}2x + 4y=-1\\24x-4y = 40\end{cases}$$

\).
For system (d): Multiply \(44x - 4y = 8\) by 10 to get \(440x-40y = 80\), resulting in the system \(

$$\begin{cases}22x + 40y=14\\440x-40y = 80\end{cases}$$

\).
(And the solution for (a) is \(\boldsymbol{(\frac{3}{2},-1)}\), the solution for (d) needs re - calculation as the hand - written one has errors.)