Solution:
Given the simultaneous equations:
[tex]\begin{gathered} 4x+3y=15\text{ --- equation 1} \\ 5x-2y=13\text{ ---- equation 2} \end{gathered}[/tex]
To solve for x and y, using the elimination method, we have
[tex]\begin{gathered} 2\times(4x+3y=15)\Rightarrow8x+6y=30\text{ --- equation 3} \\ 3\times(5x-2y=13)\Rightarrow15x-6y=39\text{ --- equation 4} \end{gathered}[/tex]
Add up equations 1 and 2.
thus, this gives
[tex]\begin{gathered} 8x+15x+6y-6y=30+39 \\ \Rightarrow23x=69 \\ divide\text{ both sides by the coefficient of x, which is 23} \\ \frac{23x}{23}=\frac{69}{23} \\ \Rightarrow x=3 \end{gathered}[/tex]
To solve for y, substitute the value of 3 for x into equation 1.
thus, from equation 1
[tex]\begin{gathered} 4x+3y=15 \\ when\text{ x = 3,} \\ 4(3)+3y=15 \\ \Rightarrow12+3y=15 \\ add\text{ -12 to both sides,} \\ -12+12+3y=-12+15 \\ 3y=3 \\ divide\text{ both sides by the coefficient of y, which is 3} \\ \frac{3y}{3}=\frac{3}{3} \\ \Rightarrow y=1 \end{gathered}[/tex]
Hence, the solution to the equation is
[tex]\begin{gathered} x=3 \\ y=1 \end{gathered}[/tex]
