Question

Find the answer by substitution method 3x+xy-7=0, 4x+y-6=0

Answer 1

No solution

Step-by-step explanation:

System of equation:

3x+xy-7=0, 4x+y-6=0

4x+y-6=0 solve for y and then substitute into first equation.

y = 6-4x

After substitution

3x+x(6-4x)-7=0

$3x+6x-4x^2-7=0$

$4x^2-9x+7=0$

factor it using quadratic formula

$x=\dfrac{9\pm\sqrt{9^2-4\cdot 4\cdot 7}}{2\cdot 4}$

$x=\dfrac{9\pm\sqrt{-31}}{8}$

x is not real solution.

So, no solution for system of equation.

#BAL

Answer 2

Step-by-step explanation:

Given Find the answer by substitution method 3x+2y-7=0, 4x+y-6=0

• Let 3x + 2y – 7 = 0 --------1
•      4x + y – 6 = 0---------2
• Given 4x + y – 6 = 0
• So y == 6 – 4x
• Substituting the value of y in equation 1 we get
• 3x + 2y – 7 = 0
• 3x + 2(6 – 4x) – 7 = 0
• 3x + 12 – 8x – 7 = 0
• 5 – 5x = 0
• Or 5x = 5
• Or x = 1
• Now substituting the value of x in equation 1 we get
• 4x + y – 6 = 0
• 4(1) + y – 6 = 0
• 4 + y – 6 = 0
• So y – 2 = 0
• Or y = 2

So the values of x and y are 1 and 2

Reference link will be

https://brainly.in/question/2271162

# BAL