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 $$-\frac{5}{3}$$

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.

$\boxed{\frac{5}{3}}$

[calc]  Question  Hard

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

What value of $$h$$ is the solution to the given equation?

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

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 ?

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?

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. $$x – 9 = 10$$
$x = 19$

2. $$x – 9 = -10$$
$x = -1$

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?

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 $2 x$ ?

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?

$.8,4 / 5$

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

C

Question

$\left&space;|&space;x-2&space;\right&space;|=3$

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

4

Questions

$x+x=9$

What value of $x$ satisfies the equation given?

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 $$b$$ ?

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. $\frac{3}{a}$

Ans: D

Questions

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

Ans: 0

Question

$\frac{1}{x-8}=-\frac{1}{x-9}$

What value of $x$ satisfies the equation above?

Ans: $8.5,17 / 2$

Questions

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

What value of $x$ satisfies the equation above?

Ans: 1/2, . 5

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