Question

5x+4y=15 and 3x+7y=12 solve by Gauss elimination method

Answer 1

Step-by-step explanation:

5x+4y=15.....(1)

3x+7y=12.....(2)

multiple eqn (1) by 3

multiple eqn (2) by 5

15x+12y=45

15x+35y=60

one is 57/23

second is 15/23

Answer 2

Answer:

The solution is $x=\frac{57}{23}$ and $y=\frac{15}{23}$.

Step-by-step explanation:

Consider the system of equations as follows:

$5x+4y=15$

$3x+7y=12$

By Gauss elimination method.

Write the system of equation in the augmented matrix form, i.e.,

$\left[\begin{array}{cc}A&b\end{array}\right]$, where $A=\left[\begin{array}{cc}5&4\\3&7\end{array}\right]$ and $b=\left[\begin{array}{c}15\\12\end{array}\right]$

⇒ $\left[\begin{array}{ccc}5&4&15\\3&7&12\end{array}\right]$

Reduced the matrix in the echelon form.

Dividing $R_1$ by $5$ to get new elements of $R_1$, i.e., $R_1 \to \frac{R_1}{5}$ as follows:

⇒ $\left[\begin{array}{ccc}1&\frac{4}{5} &3\\3&7&12\end{array}\right]$

Subtracting $3R_1$ from $R_2$ to get new elements of $R_2$, i.e., $R_2 \to R_2-3R_1$ as follows:

⇒ $\left[\begin{array}{ccc}1&\frac{4}{5} &3\\0&\ 23/5 &3\end{array}\right]$

Multiply $R_2$ by $\frac{5}{23}$ to get new elements of $R_2$ as follows:

⇒ $\left[\begin{array}{ccc}1&\frac{4}{5} &3\\0&\ 1 &\frac{15}{23} \end{array}\right]$

Here, $x+\frac{4}{5}y=3$ ...... (1)

$y=\frac{15}{23}$

Substitute the value $\frac{15}{23}$ for $y$ in the equation (1) as follows:

$x+\frac{4}{5} \cdot\frac{15}{23} =3$

⇒ $x+\frac{12}{23} =3$

⇒ $x=3-\frac{12}{23}$

⇒$x=\frac{57}{23}$

Therefore, the value of $x$ is $\frac{57}{23}$ and $y$ is $\frac{15}{23}$.

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