Home / Digital SAT Math Practice Questions – Advanced : Linear equations in one variable

Digital SAT Math Practice Questions – Advanced : Linear equations in one variable

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

[ calc]  Questions  Hard

\[
-2|x-5|=-4 x
\]

What are all possible solutions to the given equation?
A) -5
B) \(\frac{5}{3}\)
C) -5 and \(\frac{5}{3}\)
D) 5 and \(-\frac{5}{3}\)

▶️Answer/Explanation

Ans: B

To solve the equation:
\[ -2|x-5| = -4x \]

First, divide both sides by \(-2\):
\[ |x-5| = 2x \]

We consider two cases for the absolute value equation:

Case 1: \(x-5 = 2x\)
\[
x – 5 = 2x
\]
\[
-5 = x
\]
So, \(x = -5\).

Case 2: \(x-5 = -2x\)
\[
x – 5 = -2x
\]
\[
x + 2x = 5
\]
\[
3x = 5
\]
\[
x = \frac{5}{3}
\]

We must check if both solutions satisfy the original equation.

For \(x = -5\):
\[
-2| -5 – 5 | = -2| -10 | = -2 \times 10 = -20
\]
\[
-4 \times -5 = 20
\]
Clearly, \(-20 \neq 20\), so \(x = -5\) is not a valid solution.

For \(x = \frac{5}{3}\):
\[
-2 \left| \frac{5}{3} – 5 \right| = -2 \left| \frac{5}{3} – \frac{15}{3} \right| = -2 \left| \frac{-10}{3} \right| = -2 \times \frac{10}{3} = -\frac{20}{3}
\]
\[
-4 \times \frac{5}{3} = -\frac{20}{3}
\]
Both sides are equal, so \(x = \frac{5}{3}\) is a valid solution.

Thus, the correct answer is:
\[
\boxed{\frac{5}{3}}
\]

[calc]  Question  Hard

$
2.1(h+3)=3 h+2.1
$

What value of \(h\) is the solution to the given equation?

▶️Answer/Explanation

Ans: 4.66,14 / 3,4.67

 To solve the equation \(2.1(h+3) = 3h + 2.1\), we can start by distributing \(2.1\) on the left side:

\[2.1h + 2.1(3) = 3h + 2.1\]

\[2.1h + 6.3 = 3h + 2.1\]

Now, let’s isolate \(h\) by moving all terms with \(h\) to one side of the equation:

\[2.1h – 3h = 2.1 – 6.3\]

\[-0.9h = -4.2\]

Divide both sides by \(-0.9\) to solve for \(h\):

\[h = \frac{-4.2}{-0.9} = 4.67\]

So, the value of \(h\) that is the solution to the given equation is \(h = 4.67\).

[Calc]  Question  Hard

If \(\frac{x-3}{7}=\frac{x-3}{9}\) , the value of 𝑥 − 3 is between which of the following pairs of values?
A. -7 and -9
B. -1 and 1
C. 2.5 and 3.5
D. 6.75 and 9.25

▶️Answer/Explanation

Ans: B

To find the value of \(x-3\), let’s first cross multiply the equation \(\frac{x-3}{7} = \frac{x-3}{9}\):

\[ 9(x-3) = 7(x-3) \]

Expanding both sides:

\[ 9x – 27 = 7x – 21 \]

Now, let’s solve for \(x\):

\[ 9x – 7x = -21 + 27 \]
\[ 2x = 6 \]
\[ x = 3 \]

Now that we know \(x = 3\), we can substitute it back into the equation to find \(x – 3\):

\[ x – 3 = 3 – 3 = 0 \]

So, the value of \(x – 3\) is \(0\). Therefore, the correct answer is option B, \(-1\) and \(1\).

[No- Calc]  Questions   Hard

\(\frac{1}{2x}+5=kx+7\)

In the given equation, k is a constant. The equation has no solution. What is the value of k ?

▶️Answer/Explanation

Ans: 1/2, .5

To have no solution, the coefficients of \(x\) on both sides of the equation should be equal, but the constants should be different. Therefore, we set the coefficients equal to each other:

\[kx = \frac{1}{2}x \]

\[k = \frac{1}{2} \]

So, the value of \(k\) is \(\frac{1}{2}\).

[Calc]  Question   Hard

2 | x – 9 |= 20
What is the sum of the solutions to the given equation?

▶️Answer/Explanation

18

To solve the equation \( 2|x-9|=20 \), we’ll isolate \( |x-9| \) first.

Dividing both sides by 2:
\[ |x – 9| = 10 \]

This equation means that \( x – 9 \) could be either 10 or -10, since the absolute value of a number is its distance from zero on the number line.

So, we have:
\[ x – 9 = 10 \]
\[ \text{or} \]
\[ x – 9 = -10 \]

Solving each equation:
1. \( x – 9 = 10 \)
Adding 9 to both sides:
\[ x = 19 \]

2. \( x – 9 = -10 \)
Adding 9 to both sides:
\[ x = -1 \]

The sum of the solutions is \( 19 + (-1) = 18 \).

[No- Calc]  Question   Hard

$
\frac{3 x}{2}+4=13
$

What value of \(\mathrm{x}\) satisfies the given equation?

▶️Answer/Explanation

Ans:6

To solve the equation \(\frac{3x}{2} + 4 = 13\) for \(x\):

 Subtract 4 from both sides:
\[
\frac{3x}{2} + 4 – 4 = 13 – 4
\]
\[
\frac{3x}{2} = 9
\]

Multiply both sides by 2 to clear the fraction:
\[
2 \cdot \frac{3x}{2} = 9 \cdot 2
\]
\[
3x = 18
\]

Divide both sides by 3:
\[
x = \frac{18}{3}
\]
\[
x = 6
\]

The value of \(x\) is \(6\).

[Calc]  Question  Hard

If $3 x-9=6$, what is the value of $2 x$ ?

▶️Answer/Explanation

10

[Calc]  Question  Hard

If $\frac{2 x}{3}-2=\frac{x}{3}+1$, what is the value of $2 x ?

▶️Answer/Explanation

$18

[Calc]  Question Hard

$$
3 x-0.6=1.8
$$

What value of $x$ satisfies the equation above?

▶️Answer/Explanation

$.8,4 / 5$

Question

\(x\)(\(x\)+2)2=\(x\)3+b\(x\)2+c\(x\)

In the equation above, b and c are constants. If the equation is true for all values of \(x\), what is the value of b+c ? 

  1. 4
  2. 6
  3. 8
  4. 16
▶️Answer/Explanation

C

Question

\left | x-2 \right |=3

What is the sum of the solutions to the given equation? 

▶️Answer/Explanation

4

Questions 

 $x+x=9$

What value of $x$ satisfies the equation given?

▶️Answer/Explanation

Ans: 4.5, 9/2

Question 

 2\(x\)+7=\(b\)\(x\)+5

In the given equation, \(b\) is a constant. If the equation has no solution, what is the value of \(b\) ?

▶️Answer/Explanation

2

Questions 

If $a x+a=3$, where $a$ is a nonzero constant, which of the following must be equal to $x+1$ ?
A. 3
B. $\mathrm{a}$
C. $3 \mathrm{a}$
D. $\frac{3}{a}$

▶️Answer/Explanation

Ans: D

Questions 

 If $a$ is the mean and $b$ is the median of nine consecutive integers, what is the value of $|a-b|$ ?

▶️Answer/Explanation

Ans: 0

Question

 $\frac{1}{x-8}=-\frac{1}{x-9}$

What value of $x$ satisfies the equation above?

▶️Answer/Explanation

Ans: $8.5,17 / 2$

Questions 

$\frac{1}{x}+\frac{1}{x-1}=0$

What value of $x$ satisfies the equation above?

▶️Answer/Explanation

Ans: 1/2, . 5

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