Q. Solve the system of equations.y=−28x−2y=x2−14x+47Write the coordinates in exact form. Simplify all fractions and radicals.(______,______)
Substitute y: Substitute y from the first equation into the second equation. Since y=−28x−2, we can replace y in the second equation with −28x−2. The second equation becomes −28x−2=x2−14x+47.
Solve for x: Now, we need to solve for x. To do this, we will set the equation to zero by moving all terms to one side: x2−14x+47=28x+2. This simplifies to x2−42x+45=0.
Factor the quadratic equation: Next, we factor the quadratic equationx2−42x+45=0. The factors of 45 that add up to 42 are 1 and 45, but we need them to subtract to 42, so the factors are 3 and 15. This gives us (x−3)(x−15)=0.
Find solutions for x: Setting each factor equal to zero gives us the solutions for x. So, x−3=0 or x−15=0. This means x=3 or x=15.
Substitute x back: Now we substitute x back into the first equation y=−28x−2 to find the corresponding y values. First, for x=3, we get y=−28(3)−2, which simplifies to y=−84−2, so y=−86.
Find corresponding y values: Next, for x=15, we substitute into the first equation y=−28x−2 to get y=−28(15)−2, which simplifies to y=−420−2, so y=−422.
Identify solutions: We now have two solutions for the system of equations: (3,−86) and (15,−422). These are the coordinates of the points where the two equations intersect.
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