Question

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

Answer 1

Step-by-step explanation:

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

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

• from eq (1)

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

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

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

• Substitute the value in eq(2)

${:}\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={\dfrac {77}{11}}$

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

• Substitute the value in eq (3)

${:}\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)}}}$