# = 3x + 42 2 in .NET framework Creation Data Matrix ECC200 in .NET framework = 3x + 42 2

216 = 3x + 42 2
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Then subtract 42 from both sides and divide both sides by 3, giving the desired result 108 42 = 3x 108 42 =x 3 x = 22 3 Answer: C The equation is in the form of a quadratic equation ax2 + b x + c = 0, where a = 1, b = 3, and c = 1 To solve this problem, you use the quadratic formula or b b2 4ac 3 x= = 2a
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( ) ( 3)2 4(1)(1) = 3 2(1)
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9 4 3 5 = 2 2
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PART VI: REVIEWING PCAT MATH SKILLS
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The roots of the quadratic equation are: x= 4 Answer: B Write the equations as (1) x 3y = 7 (2) 4x + 3y = 2 Solve Equation (1) for x in terms of y or (3) x=3y 7 3+ 5 3 5 , 2 2
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Substitute Equation (3) into Equation (2) to yield: 4(3y 7) + 3y = 2 12y 28 + 3y = 2 15y = 30 30 y= = 2 15 Substituting this value for y into Equation (1) yields the value for x: x 3 (2) = 7 x= 7+6= 1 So the solution for the system of equations is x = 1, y = 2 5 Answer: D From the rst equation, multiply both sides by y resulting in x = 5y Because x = 25, you can write 25 = 5y 25 = 5 or y = 5 Substituting the given information regarding x and y into its sum yields: x + y = 25 + 5 = 30 6 Answer: A To determine the quotient of 6x3 y4 z2 divided by 2xy3 z5, write the two monomials as a division problem with one expression on top of another: 6x3y 4 z2 2xy3z5 The expressions are then arranged and presented such that you can calculate the quotients of the coef cients and of the variables:
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6x3y 4 z2 6 x3 y 4 z2 3x2 y = = 3 x2 y z 3 = 2xy3z5 2 x y3 z5 z3 7 Answer: C x2 4x 21 = 1, you must rst try to simplify the numerator by Given the equation x+3 expressing the polynomial in terms of its factors or x2 4x 21 x 7 x + 3 =1 = x+3 x+3
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CHAP 13: ALGEBRA
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The term (x + 3) can be eliminated from the numerator and denominator because any term divided by itself is 1 Thus, the equation above can be simpli ed as: (x 7) = 1 or x = 8 8 Answer: B To determine the roots of a quadratic equation, you can rst try to see if the equation can be factored In this case, the equation 2x 2 6x 20 = 0 can be factored as: 2x 2 6x 20 = (2x + 4)(x 5) = 0 Thus, the roots can be found by the two equations: 2x + 4 = 0 x 5=0 which yields the root x = 2 which yields the root x = 5
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9 Answer: C When asked to solve an absolute value equation, it is helpful to write the equation in its two forms: (1) 2x 4 = 20 (2) 2x 4 = 20 Solving for x in equation (1) yields x = 12 and in equation (2) yields x = 8 10 Answer: A In order to determine the value of x from the expression, you must perform mathematical operations to isolate x The rst step is to undo the cube root by raising each side of the equation by an exponent equal to 3 or 27 3 3 x 4 = 3
27 = 27 x 4 27 = 27 (x 4) 27 = 27 x 108 27 + 108 = 27x x = 5
CRAM SESSION
Algebra
An algebraic expression is a collection of ordinary numbers, letters, and operational signs Examples include 3t2 , 4x 2 + 2y , 6a 4 8b 3 10c2 , and 7(4t s) Each individual collection of numbers and letters is referred to as a term An algebraic expression with one term is called a monomial An algebraic expression with more than one term is called a polynomial A polynomial with two terms is a binomial A polynomial with three terms is a trinomial