Question

what is the point of intersection when the system of equations below is graphed on the coordinate plane -x+y=4 6x+y=-3​

Answer 1

Answer:

POINTS OF INTERSECTION ARE:

x = -1

y = 3

coordinates : (-1,3)

Coordinates :

Explanation:

Condition for intersecting lines

$\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}$

here

a1 ,a2 = coefficients of x (a1 = -1 , a2 = 6)

b1 ,b2 = coefficients of y(b1 = 1 , b2 = 1)

$frac{-1}{6}\neq\frac{1}{1}\neq\frac{1}{1}$

Calculation :

Here substitution method is used : Value of y from eq (1) is substitued in equation (2)

-x + y = 4

y = 4 + x.............(i)

put value of x in equation (2)

6x + y = -3

6x + 4 + x = -3  ( from i)

7x = -3 - 4

7x = -7

x = $frac{-7}{7}$

x = -1

using equation(i)

y = 4 + (-1)

y = 4 - 1

y = 3

Check:

The values of x and y should satisfy equation (1) and (2)

for equation 1

-x+y =4

= -(-1) + 3

= 1 + 3

= 4

Hence coordinates satisfy equation 1