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DSAT Math-Systems of two linear equations in two variables- Practice Questions

DSAT Math-Systems of two linear equations in two variables- Practice Questions

SAT MAth Practice questions – all topics

  • Algebra Weightage: 35%  Questions: 13-15
    • Linear equations in one variable
    • Linear equations in two variables
    • Linear functions
    • Systems of two linear equations in two variables
    • Linear inequalities in one or two variables

SAT MAth and English  – full syllabus practice tests

Question   Easy

\[
\begin{aligned}
& y<x \\
& x>7
\end{aligned}
\]

What point \((x, y)\) is a solution to the given system of inequalities in the \(x y\) plane?
A) \((0,1)\)
B) \((3,2)\)
C) \((6,7)\)
D) \((9,8)\)

▶️Answer/Explanation

Ans:D

The given system of inequalities is:

\[
\begin{aligned}
& y < x \\
& x > 7
\end{aligned}
\]

We need to find the point \((x, y)\) that satisfies both inequalities.

Let’s evaluate the given options:

A) \((0, 1)\)
\(0 > 7\) (False)
\(1 < 0\) (False)
This point does not satisfy \(x > 7\).

B) \((3, 2)\)
\(3 > 7\) (False)
\(2 < 3\) (True)
This point does not satisfy \(x > 7\).

C) \((6, 7)\)
\(6 > 7\) (False)
\(7 < 6\) (False)
This point does not satisfy \(x > 7\).

D) \((9, 8)\)
\(9 > 7\) (True)
\(8 < 9\) (True)
This point satisfies both \(x > 7\) and \(y < x\).

  Question Easy

y=18x+25
y =- 14x-7
What is the solution (x,y) to the given system of equations?

A. (-7,25)

B. (-1,7)

C. (7,-1)

D. (25,-7)

▶️Answer/Explanation

Ans: B

To solve this system, we set the equations equal to each other since they both equal \(y\):
\[18x + 25 = -14x – 7\]

Now, let’s solve for \(x\):
\[18x + 14x = -7 – 25\]
\[32x = -32\]
\[x = -1\]

Now that we have \(x = -1\), let’s find \(y\) by substituting \(x\) into one of the original equations. We’ll use the first equation:
\[y = 18(-1) + 25\]
\[y = -18 + 25\]
\[y = 7\]

So, the solution \((x, y)\) to the given system of equations is \((-1, 7)\), which corresponds to option B).

  Question Easy

\[
\begin{aligned}
& x=4 \\
& y=\frac{x}{4}+2
\end{aligned}
\]

What is the solution \((x, y)\) to the given system of equations?
A) \((4,6)\)
B) \((4,3)\)
C) \((4,2)\)
D) \((4,1)\)

▶️Answer/Explanation

B

We need to find the solution \((x, y)\) to the given system of equations.

First, substitute \(x = 4\) into the equation for \(y\):
\[ y = \frac{4}{4} + 2 \]
\[ y = 1 + 2 \]
\[ y = 3 \]

Therefore, the solution to the system of equations is:
\[ (x, y) = (4, 3) \]

So the answer is:
\[ \boxed{B} \]

 Question  Easy

The table shows the prices of 3 items in a certain store on January \(15,1913\).

On January 15,1913 , Samuel purchased \(s\) pounds of sugar and \(p\) pounds of potatoes for a total of \(\$ 0.16\). The total weight of the purchase was 4 pounds. Based on the prices in the table, which system of equations represents this situation?

A) \(0.06 s+0.02 p=0.16\)
\(s+p=4\)

B) \(6 s+2 p=0.16\)
\(s+p=4\)

C) \(0.06 s+0.02 \mathrm{p}=4\)
\(s+p=0.16\)

D) \(6 s+2 p=4\)
\(s+p=0.16\)

▶️Answer/Explanation

Ans: A

The total cost of the purchase was \(\$ 0.16\). Therefore, the equation for the cost is:
\[
0.06 s+0.02 p=0.16
\]

The total weight of the purchase was 4 pounds. Therefore, the equation for the weight is:
\[
s+p=4
\]

Putting these two equations together, the system of equations representing this situation is:
\[
\left\{\begin{array}{l}
0.06 s+0.02 p=0.16 \\
s+p=4
\end{array}\right.
\]

  Question   Foundation

$
\begin{aligned}
& x=3 \\
& y=x+3
\end{aligned}
$

What is the solution \((x, y)\) to the given system of equations?

A) \((3,6)\)
B) \((3,3)\)
C) \((3,-3)\)
D) \((3,-6)\)

▶️Answer/Explanation

Ans:A

1. From the first equation, we know:
\[
x = 3
\]

2. Substitute \(x = 3\) into the second equation:
\[
y = 3 + 3 = 6
\]

So, the solution to the system of equations is \((x, y) = (3, 6)\).

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