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Welcome to today’s tutorial! In this post, we break down some of the top SAT Mathematics questions, offering clear step-by-step solutions. These problems focus on algebra, trigonometry, and quadratic equations—core areas tested in every SAT exam.


You can watch the full class in the video below:


Question 1: Solving Simultaneous Equations

Given: 3x+2y=12(1)2x−y=4(2)3x + 2y = 12 \quad \text{(1)} \\ 2x – y = 4 \quad \text{(2)}3x+2y=12(1)2x−y=4(2)

Find: x+yx + yx+y

Step-by-Step Solution:

We’ll use the elimination method to solve for xxx and yyy.

First, multiply equation (2) by 2: 4x−2y=8(3)4x – 2y = 8 \quad \text{(3)}4x−2y=8(3)

Now add equation (1) and equation (3): (3x+2y)+(4x−2y)=12+8⇒7x=20⇒x=207(3x + 2y) + (4x – 2y) = 12 + 8 \Rightarrow 7x = 20 \Rightarrow x = \frac{20}{7}(3x+2y)+(4x−2y)=12+8⇒7x=20⇒x=720​

Substitute x=207x = \frac{20}{7}x=720​ back into equation (1): 3(207)+2y=12⇒607+2y=12⇒2y=247⇒y=1273\left(\frac{20}{7}\right) + 2y = 12 \Rightarrow \frac{60}{7} + 2y = 12 \Rightarrow 2y = \frac{24}{7} \Rightarrow y = \frac{12}{7}3(720​)+2y=12⇒760​+2y=12⇒2y=724​⇒y=712​

Now add x+yx + yx+y: 207+127=327≈4.57\frac{20}{7} + \frac{12}{7} = \frac{32}{7} \approx 4.57720​+712​=732​≈4.57

Answer: Approximately 5


Question 2: Identifying a Quadratic Equation from Roots

Given Roots:
x=−2x = -2x=−2 and x=5x = 5x=5

Solution:

From the roots, we form: (x+2)(x−5)⇒f(x)=(x+2)(x−5)(x + 2)(x – 5) \Rightarrow f(x) = (x + 2)(x – 5)(x+2)(x−5)⇒f(x)=(x+2)(x−5)

This is the required quadratic function.

Answer: f(x)=(x+2)(x−5)f(x) = (x + 2)(x – 5)f(x)=(x+2)(x−5)


Question 3: Trigonometry in Right Triangles

Given: sin⁡(θ)=35,θ in the first quadrant\sin(\theta) = \frac{3}{5}, \quad \theta \text{ in the first quadrant}sin(θ)=53​,θ in the first quadrant

Solution:

Use the identity: sin⁡2(θ)+cos⁡2(θ)=1⇒(35)2+cos⁡2(θ)=1⇒925+cos⁡2(θ)=1⇒cos⁡2(θ)=1625⇒cos⁡(θ)=45\sin^2(\theta) + \cos^2(\theta) = 1 \Rightarrow \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 \Rightarrow \frac{9}{25} + \cos^2(\theta) = 1 \Rightarrow \cos^2(\theta) = \frac{16}{25} \Rightarrow \cos(\theta) = \frac{4}{5}sin2(θ)+cos2(θ)=1⇒(53​)2+cos2(θ)=1⇒259​+cos2(θ)=1⇒cos2(θ)=2516​⇒cos(θ)=54​

Answer: cos⁡(θ)=45\cos(\theta) = \frac{4}{5}cos(θ)=54​


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