Question

nom:
4 cst chapitre 4 : devoir #2
1 résous les systèmes d’équations suivants à l’aide de la méthode de comparaison.
a)
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$$\begin{cases} y = 4x + 5 \\\\ y = \\dfrac{8 + 5x}{3} \\end{cases}$$

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b)
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$$\begin{cases} y = \\dfrac{x}{2} + 4 \\\\ y = 2x - 11 \\end{cases}$$

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2 résous les systèmes d’équations suivants à l’aide de la méthode de substitution.
a)
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$$\begin{cases} 2x + 4y + 40 = 0 \\\\ 7x + 5 = y \\end{cases}$$

\\)
b)
\\(\

$$\begin{cases} 3x + 5y = 51 \\\\ x = 3y + 10 \\end{cases}$$

\\)

Explanation:

Problem 1a (Method of Comparison)

Step 1: Set the two equations equal

Since both equal \( y \), set \( 4x + 5=\frac{8 + 5x}{3} \)

Step 2: Multiply both sides by 3

\( 3(4x + 5)=8 + 5x \)
\( 12x + 15 = 8 + 5x \)

Step 3: Subtract \( 5x \) and 15

\( 12x - 5x=8 - 15 \)
\( 7x=-7 \)

Step 4: Solve for \( x \)

\( x = -1 \)

Step 5: Substitute \( x \) into \( y = 4x + 5 \)

\( y = 4(-1)+5 = 1 \)

Step 1: Set the two equations equal

\( \frac{x}{2}+4 = 2x - 11 \)

Step 2: Multiply both sides by 2

\( x + 8 = 4x - 22 \)

Step 3: Subtract \( x \) and add 22

\( 8 + 22 = 4x - x \)
\( 30 = 3x \)

Step 4: Solve for \( x \)

\( x = 10 \)

Step 5: Substitute \( x \) into \( y = 2x - 11 \)

\( y = 2(10)-11 = 9 \)

Step 1: Substitute \( y = 7x + 5 \) into \( 2x + 4y + 40 = 0 \)

\( 2x + 4(7x + 5)+40 = 0 \)

Step 2: Expand and simplify

\( 2x + 28x + 20 + 40 = 0 \)
\( 30x + 60 = 0 \)

Step 3: Solve for \( x \)

\( 30x=-60 \)
\( x=-2 \)

Step 4: Substitute \( x \) into \( y = 7x + 5 \)

\( y = 7(-2)+5=-9 \)

Answer:

\( x=-1,y = 1 \)

Problem 1b (Method of Comparison)