Question

h51a
kumon
h 51
simultaneous equations
with two variables 4
grade — — c d
solve the following simultaneous equations.
ex.
\\\

$$\begin{cases} \\frac{x + 1}{2} = \\frac{y}{3} \\dots \\textcircled{1} \\\\ \\frac{x}{3} = y - 5 \\dots \\textcircled{2} \\end{cases}$$

\\
sol
\textcircled{1}\times6 \\( 3(x + 1) = 2y \\)
\\( 3x - 2y = -3 \dots \textcircled{3} \\)
\textcircled{2}\times3 \\( x = 3(y - 5) \\)
\\( x - 3y = -15 \dots \textcircled{4} \\)
\textcircled{4}\times3 \\( 3x - 9y = -45 \dots \textcircled{5} \\)
\textcircled{3}\\u2212\textcircled{5} \\( 7y = 42 \\)
\\( y = 6 \\)
substituting \\( y = 6 \\) into \textcircled{4},
\\( x - 18 = -15 \\)
\\( x = 3 \\)
\\\

$$\begin{cases} x = 3 \\ y = 6 \\end{cases}$$

\\
(verification)
substituting \\( x = 3 \\) and \\( y = 6 \\) into
\textcircled{1} and \textcircled{2},
\textcircled{1}
\\\

$$\begin{cases} \\text{lhs} = \\frac{3 + 1}{2} = 2 \\\\ \\text{rhs} = \\frac{6}{3} = 2 \\end{cases}$$

\\
\textcircled{2}
\\\

$$\begin{cases} \\text{lhs} = \\frac{3}{3} = 1 \\\\ \\text{rhs} = 6 - 5 = 1 \\end{cases}$$

\\
(1)
\\\

$$\begin{cases} \\frac{x - 1}{3} = \\frac{y}{4} \\\\ \\frac{x}{2} = y + 3 \\end{cases}$$

\\
sol
(verification)
\textcircled{1}
\\\

$$\begin{cases} \\text{lhs} = \\\\ \\text{rhs} = \\end{cases}$$

\\
\textcircled{2}
\\\

$$\begin{cases} \\text{lhs} = \\\\ \\text{rhs} = \\end{cases}$$

\\

Explanation:

Step1: Simplify the first equation

Multiply both sides of \(\frac{x - 1}{3}=\frac{y}{4}\) by \(12\) (the least common multiple of \(3\) and \(4\)) to eliminate denominators.
\[

$$\begin{align*} 12\times\frac{x - 1}{3}&=12\times\frac{y}{4}\\ 4(x - 1)&=3y\\ 4x-4&=3y\\ 4x - 3y&=4\quad\cdots③ \end{align*}$$

\]

Step2: Simplify the second equation

Multiply both sides of \(\frac{x}{2}=y + 3\) by \(2\) to simplify.
\[

$$\begin{align*} 2\times\frac{x}{2}&=2\times(y + 3)\\ x&=2y+6\\ x - 2y&=6\quad\cdots④ \end{align*}$$

\]

Step3: Eliminate \(x\) to solve for \(y\)

Multiply equation ④ by \(4\) to make the coefficients of \(x\) in both equations equal.
\[

$$\begin{align*} 4\times(x - 2y)&=4\times6\\ 4x-8y&=24\quad\cdots⑤ \end{align*}$$

\]
Subtract equation ⑤ from equation ③:
\[

$$\begin{align*} (4x - 3y)-(4x - 8y)&=4 - 24\\ 4x - 3y - 4x + 8y&=-20\\ 5y&=-20\\ y&=- 4 \end{align*}$$

\]

Step4: Substitute \(y\) to solve for \(x\)

Substitute \(y = - 4\) into equation ④ (\(x-2y = 6\)):
\[

$$\begin{align*} x-2\times(-4)&=6\\ x + 8&=6\\ x&=6 - 8\\ x&=-2 \end{align*}$$

\]

Step5: Verification

For equation ① \(\frac{x - 1}{3}=\frac{y}{4}\)

Substitute \(x=-2\) and \(y = - 4\):

• Left - Hand Side (LHS): \(\frac{-2 - 1}{3}=\frac{-3}{3}=-1\)
• Right - Hand Side (RHS): \(\frac{-4}{4}=-1\)

For equation ② \(\frac{x}{2}=y + 3\)

Substitute \(x=-2\) and \(y = - 4\):

• Left - Hand Side (LHS): \(\frac{-2}{2}=-1\)
• Right - Hand Side (RHS): \(-4 + 3=-1\)

Answer:

\(

$$\begin{cases}x=-2\\y = - 4\end{cases}$$

\)