Evaluate the expression. (26÷13)⋅(−7)+14−4
2 answers:
You might be interested in
Part A.
The set of equations can be solved by substitution. Use the expression one equation gives for y as the value of y in the other equation. This gives
2x²-15 = 3x-6
Subtracting the right side gives a quadratic in standard form that can be solved by any of several methods.
2x² -3x -9 = 0
(2x+3)(x-3) = 0 . . . . factor the above equation
x = -3/2, x = 3 . . . . .use the zero product rule to find x
Now, these x-values can be substituted into either equation for y. The linear equation is often easier to evaluate.
y = 3(-3/2) -6 = -10.5
y = 3(3)-6 = 3
The solutions to the system are (-1.5, -10.5) and (3, 3).
Part B.
The two equations can be graphed. The solutions are where the graphs intersect. The graphs intersect where the (x, y) values that satisfy one equation are the same (x, y) values that satisfy the other equation. Those points of intersection are (-1.5, -10.5) and (3, 3).
The set of equations can be solved by substitution. Use the expression one equation gives for y as the value of y in the other equation. This gives
2x²-15 = 3x-6
Subtracting the right side gives a quadratic in standard form that can be solved by any of several methods.
2x² -3x -9 = 0
(2x+3)(x-3) = 0 . . . . factor the above equation
x = -3/2, x = 3 . . . . .use the zero product rule to find x
Now, these x-values can be substituted into either equation for y. The linear equation is often easier to evaluate.
y = 3(-3/2) -6 = -10.5
y = 3(3)-6 = 3
The solutions to the system are (-1.5, -10.5) and (3, 3).
Part B.
The two equations can be graphed. The solutions are where the graphs intersect. The graphs intersect where the (x, y) values that satisfy one equation are the same (x, y) values that satisfy the other equation. Those points of intersection are (-1.5, -10.5) and (3, 3).