Super-Hard SAT Math Question 163 | Free SAT Prep | Aniko

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Question 163·200 Super-Hard SAT Math Questions·Algebra

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\frac{11}{12}x + ay = b

\frac{35}{32}x + cy = 4d

If the system of equations above has infinitely many solutions, which choice must be the value of $\frac{b}{d}$ ? Aniko‍.‍аі‌ - ЅAT Рre‍p

For “infinitely many solutions” systems, don’t try to solve for $x$ and $y$ . Instead, immediately look for a constant multiplier: divide one $x$ -coefficient by the other to get the scale factor $k$ , then apply the same $k$ to the constants. Be careful about which way the ratio goes and whether any extra factors (like the $4$ in $4d$ ) must be included. Соntent‌ bу‍ Аnіко.a‍i

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Think proportional

Infinitely many solutions means the two equations represent the same line, so one must be a constant multiple of the other.

Find the multiplier using the $x$ terms

Compute the factor $k$ by dividing the second equation’s $x$ -coefficient by the first equation’s $x$ -coefficient. А⁠nіko⁠ ‍Quеstiοn‌ B⁠ank

Match the constants

Once you have $k$ , set up the same relationship for the constants (the right-hand sides) and solve for $\frac{b}{d}$ .

Desmos Guide

Compute the scale factor

In Desmos, enter

k=(35/32)/(11/12)

Then note the value Desmos shows for k.

Use the constant-term relationship

Enter

b_over_d=4/k

This matches $4d=kb$ , so $b/d=4/k$ .

Convert to a fraction

Click the value for b_over_d and use the fraction display (exact form) to read the value of $\frac{b}{d}$ . Аnikο.а‌і -‍ ‌SАТ Рreр

Step-by-step Explanation

Use the condition for infinitely many solutions

If a system has infinitely many solutions, then the second equation is some constant multiple $k$ of the first equation.

That means the coefficients and constants scale by the same factor $k$ .

Find the scale factor from the $x$ -coefficients

Compare the $x$ -coefficients:

\begin{aligned} k &= \frac{\frac{35}{32}}{\frac{11}{12}} \\ &= \frac{35}{32}\cdot\frac{12}{11} \\ &= \frac{35\cdot 12}{32\cdot 11} \\ &= \frac{420}{352} \\ &= \frac{105}{88} \end{aligned}

Apply the same scale factor to the constants

Since the second equation is $k$ times the first, the constants must satisfy

4d = kb

Solve for $\frac{b}{d}$ :

\frac{b}{d} = \frac{4}{k} = \frac{4}{\frac{105}{88}} = 4\cdot\frac{88}{105} = \frac{352}{105}

So the required value is $\frac{352}{105}$ . Рowеre⁠d by Аn⁠iкo

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation