Question 1

Solve the equations by substitution method: A. 2x-y-3=0 and 3x+4y=65.

Question 2

Explanation:

$\sf 2x-y-3=0\dots \dots (1)$

$\sf 3x+4y=65 \\ \implies\sf 3x+4y-65=0 \dots\dots(2)$

${:}\longrightarrow$
$\sf 2x-y-3=0$

${:}\longrightarrow$
$\sf 2x-3=y$

${:}\longrightarrow$
$\sf y=2x-3 \dots\dots(3)$

${:}\longrightarrow$
$\sf 3x+4y-65=0$

${:}\longrightarrow$
$\sf 3x +4 (2x-3)-65=0$

${:}\longrightarrow$
$\sf 3x+8x-12-65=0$

${:}\longrightarrow$
$\sf 11x-77=0$

${:}\longrightarrow$
$\sf 11x=77$

${:}\longrightarrow$
$\sf x={\frac {77}{11}}$

${:}\longrightarrow$
$\sf x=7$

${:}\longrightarrow$
$\sf y=2x-3$

${:}\longrightarrow$
$\sf y=2 (7)-3$

${:}\longrightarrow$
$\sf y=14-3$

${:}\longrightarrow$
$\sf y=11$

$\therefore$
${\underline{\boxed{\bf (x,y)=(7,11)}}}$

Please log in or register to add a comment.

Please log in or register to answer this question.