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Method: Solving a System of a Linear Equation and a Hyperbolic Equation
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1 Solve for y in the linear equation and then substitute the x side of it into the y place of the hyperbolic equation 2 Rearrange the terms in order to simplify this equation into a quadratic equation 3 Solve the quadratic equation for x, either by factoring or with the Quadratic Formula 4 Substitute this value into one of the original equations to find the value of y 5 Check whether the problem is asking for two solutions or just for the positive solution (Keep in mind that there may only be one solution)
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1 If y = x + 2 and xy = 6 , what are the possible values of x and y 2
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Solution Since y is already isolated in the linear equation, substitute the right-hand side of the first
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equation into the y place of the second equation and simplify: xy = 6 x 1 x + 2 = 6 1 x 2 + 2x = 6 2 2
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54# THE QUANTITATIVE SECTION
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Then, rearrange the terms to find the resulting quadratic equation: 1 x 2 + 2x 6 = 0 2 Multiply both sides of the equation by 2 This will facilitate factoring the equation (remember, GMAT quadratics are generally factorable) x 2 + 4x 12 = 0 Factor the quadratic:
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( x + 6 )( x 2 ) = 0
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which yields the solutions x = 6 or 2 Substitute these values one at a time into the second equation to find two values of y : xy = 6 6 y = 6 y = 1 xy = 6 2y = 6 y = 3 So, the solutions to the system are ( 6, 1) and (2, 3)
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APPLICATION PROBLEM: DEPRECIATION
At this point, you should try to apply these principles to another practical problem The following problem is of a type sometimes encountered on the GMAT; it tests your ability to identify variables, construct systems of linear equations, and solve those equations with basic algebraic techniques It is challenging by the standards of the test, but it is definitely representative of typical problems that you will encounter in an MBA program, particularly in an introductory course on financial accounting Ace Transport has acquired two trucks to carry different types of cargo One costs $70,000; the other costs $52,000 The company expects that the first will last 165,000 miles, after which it can be sold for salvage for about $4,000 The other will fetch the same salvage price, but will last 200,000 miles The two trucks convoy together, so they always show the same mileage readings If the value of each truck declines at different constant rates for every mile driven, at what mileage reading will the residual values of the two trucks be equal A 87,500 miles B 100,000 miles C 107,750 miles D 112,500 miles E The residual values will never be equal
Solution We can take the value of the truck to depend on the reading on the odometer, which makes
the mileage the independent variable and the value of the truck the dependent variable The clue that the values of the trucks decline by an equal amount every mile indicates that this is what accountants
ALGEBRAIC EQUATIONS AND ANALYTICAL GEOMETRY $55
call a straight-line depreciation problem That is, the rate of decrease is constant, so the slope is constant We can determine rather quickly the respective rates of decrease, or depreciation, using the equation r = vi v m
where v is the current value of the truck; v-sub-i is the initial value, or price, and m is the reading on the odometer At the final point of usage, this is r = vi v f mf
where v-sub-f is the final, or salvage, value, which in both cases is $4,000; and m-sub-f is the final reading on the odometer For trucks 1 and 2, these work out to be r1 = $70, 000 $4, 000 = $040/mile 165, 000 miles n and r2 = $52, 000 $4, 000 = $024/mile 200, 000 miles
and since these are rates of depreciation, they are both negative Thus, r1 = $040/mile and r2 = 024/mile Note: These equations are just using the formula for slope, y2 y1/x2 x1, where for the first truck, (x1,y1) = (0, 70,000) and (x2,y2) = (165,000, 4,000) (since at 0 miles the truck s value is $70,000, and at 165,000 miles the truck s value is $4,000) Therefore, r1 = 70, 000 4, 000 66, 000 4 = = 04 dollar per mile 165, 000 0 165, 000
For the second truck, (x1,y1) = (0, 52,000) and (x2,y2) = (200,000, 4,000) (since at 0 miles the truck s value is $52,000 and at 200,000 miles the truck s value is $4000) Therefore, r1 = 52, 000 4, 000 48, 000 2 = = 024 dollar per mile 200, 000 0 200, 000
Next, we can graph these declines, remembering that depreciations are negative effects on value, so the slopes will be 040 and 024 dollar/mile (This is shown in the figure that follows) From the graph, we can tell that the answer lies somewhere around 100,000 miles While we re not sure exactly where, we can at least rule out answer choice E However, we can determine this exactly by equating the values To do that, we need to set up the equations with the current value terms exposed: r = vi v rm = vi v v = rm + vi m
Now we have a general linear equation with a slope (the depreciation rate) and a y intercept (the price paid) (Note also that our slope is r, not m; we have been using m as the variable for slope, but really any letter will do In this problem, we chose to use m as the number of miles) We can establish a system of two equations by plugging in the values from the original question: v = ( 040 ) m + 70, 000
v = ( 024 ) m + 52, 000
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