Given equation(s):
To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.
Given value of the y-variable: 3x - 1
Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.
[tex]\implies 3x + y = -7[/tex]
[tex]\implies3x + (3x - 1) = -7[/tex]
[tex]\implies3x + 3x - 1 = -7[/tex]
Combine like terms as needed;
[tex]\implies 3x + 3x - 1 = -7[/tex]
[tex]\implies 6x - 1 = -7[/tex]
Add 1 to both sides of the equation;
[tex]\implies 6x - 1 + 1 = -7 + 1[/tex]
[tex]\implies 6x = -6[/tex]
Divide 6 to both sides of the equation;
[tex]\implies \dfrac{6x}{6} = \dfrac{-6}{6}[/tex]
[tex]\implies x = -1[/tex]
Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.
[tex]\implies y = 3(-1) - 1[/tex]
[tex]\implies[/tex] [tex]\ \ = -3 - 1[/tex]
[tex]\implies[/tex] [tex]= -4[/tex]
Therefore, the value(s) of the x-variable and the y-variable are;
[tex]\boxed{x = -1}[/tex] [tex]\boxed{y = -4}[/tex]
Convert the equations into slope intercept form;
[tex]\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.[/tex]
[tex]\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.[/tex]
Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.
[tex]\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.[/tex]
[tex]\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.[/tex]
Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.
[tex]\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.[/tex]
[tex]\implies 0 = (6x) + (6)[/tex]
[tex]\implies0 = 6x + 6[/tex]
This problem is now an algebraic problem. Isolate "x" to determine its value.
[tex]\implies 0 - 6 = 6x + 6 - 6[/tex]
[tex]\implies -6 = 6x[/tex]
[tex]\implies -1 = x[/tex]
Like done in method 1, substitute the value of x into the first equation to determine the value of y.
[tex]\implies y = 3(-1) - 1[/tex]
[tex]\implies y = -3 - 1[/tex]
[tex]\implies y = -4[/tex]
Therefore, the value(s) of the x-variable and the y-variable are;
[tex]\boxed{x = -1}[/tex] [tex]\boxed{y = -4}[/tex]
The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.
⇒ (x, y) ⇒ (-1, -4)
Therefore, the point of intersection is (-1, -4).
