1) -3(5x+2y=-3)⇒ -15x-6y=9
⇒ -9x=27
2(3x+3y=9)⇒ 6x+6y=18
2) -9x/-9=27/-9 ⇒ x=-3
3) 3(-3)+3y=9⇒ -9+9+3y=9+9⇒ 3y/3=18/3⇒ y=6
Answer: (-3,6)
Reasoning:
Step 1) In order to eliminate, first I had to multiple the first equation by -3 and the second by 2 so that when combining the equations y would cancel each other out so that I could solve for x. <em>Note: There are many combinations as to how you could multiple the equations so that either the x or y would cancel out.
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Step 2) Once y is eliminated, solve for x.
Step 3) Now plug x back into one of the original equations and solve for y. <em>Note: Plug x back into one of the original equations, not the equations that were changed by multiplication,</em>
Answer:
The answer is (3x^2-2x+5)
We have a system of equations in two variables, namely, a and c. Use the substitution to find a and c.
Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60
Answer:
What are the two equations?
Step-by-step explanation:
Once u answer that I'll edit this and answer ur question :)