Solve the simultaneous equations:

[tex]
\begin{array}{l}
2x + 4y = 1 \\
3x - 5y = 7
\end{array}
[/tex]

Optional working:

[tex]x = \square[/tex]

[tex]y = \square[/tex]



Answer :

To solve the simultaneous equations:
[tex]\[ \begin{array}{l} 2x + 4y = 1 \\ 3x - 5y = 7 \end{array} \][/tex]

we can use the method of elimination or substitution to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Here, I will outline the solution step-by-step using the elimination method.

1. Write down the equations:

[tex]\[2x + 4y = 1 \tag{1}\][/tex]
[tex]\[3x - 5y = 7 \tag{2}\][/tex]

2. Multiply the equations to align coefficients for elimination:

To eliminate [tex]\(y\)[/tex], we can multiply Equation (1) by 5 and Equation (2) by 4:

[tex]\[\begin{aligned} &5 \cdot (2x + 4y) = 5 \cdot 1 \\ &4 \cdot (3x - 5y) = 4 \cdot 7 \\ \end{aligned} \][/tex]

This results in:

[tex]\[\begin{aligned} &10x + 20y = 5 \tag{3} \\ &12x - 20y = 28 \tag{4} \end{aligned} \][/tex]

3. Add the new equations to eliminate [tex]\(y\)[/tex]:

[tex]\[\begin{aligned} (10x + 20y) + (12x - 20y) &= 5 + 28 \\ 10x + 12x &= 33 \\ 22x &= 33 \\ x &= \frac{33}{22} \\ x &= \frac{3}{2} \\ x &= 1.5 \end{aligned}\][/tex]

4. Substitute [tex]\(x = 1.5\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex]:

Using Equation (1):

[tex]\[\begin{aligned} 2(1.5) + 4y &= 1 \\ 3 + 4y &= 1 \\ 4y &= 1 - 3 \\ 4y &= -2 \\ y &= \frac{-2}{4} \\ y &= -\frac{1}{2} \\ y &= -0.5 \end{aligned}\][/tex]

Therefore, the solution to the system of equations is:

[tex]\[ x = 1.5, \][/tex]
[tex]\[ y = -0.5. \][/tex]