Home / Digital SAT Math Practice Questions – Advanced : Systems of two linear equations in two variables

Digital SAT Math Practice Questions – Advanced : Systems of two linear equations in two variables

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 Hard

\[
\begin{aligned}
& -8 x-24=10 y \\
& 15 y=6-18 x
\end{aligned}
\]

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

▶️Answer/Explanation

Ans: 7

To solve the system of equations:
\[
\begin{cases}
-8x – 24 = 10y \\
15y = 6 – 18x
\end{cases}
\]

We can first isolate \(y\) in terms of \(x\) from the second equation and divide by 3 both numerator and denominator:
\[
15y = 6 – 18x \\
y = \frac{6 – 18x}{15} = \frac{2 – 6x}{5}
\]

Now, we can substitute this expression for \(y\) into the first equation:
\[
-8x – 24 = 10\left(\frac{2 – 6x}{5}\right)
\]
\[
-8x – 24 = 4 – 12x
\]

Solve for \(x\):
\[
-8x + 12x = 2 + 24
\]
\[
4x = 28
\]
\[
x = \frac{28}{4}
\]
\[
x = 7
\]

Question   Hard

2x-2y=2

One of the two linear equations in a system is given. The system has exactly one solution. Which equation could be the second equation in this system?

A. -8x+8y =- 3

B. 3x-3y=8

C. -10x+8y=5

D. -x+y =- 1

▶️Answer/Explanation

Ans: C

To find the second equation that forms a system with the given equation $2x – 2y = 2……(1)$, and has exactly one solution, we need to find an equation that is not a multiple of the given equation and intersects it at a single point.

A. -8x + 8y = -3
If we multiply the given equation(1) by -4, we get $-8x+8y = -8$.
This equation is parallel to $-8x + 8y = -3$, which means they have the same slope but different y-intercepts.
Parallel lines do not intersect, so this option is incorrect.

B. 3x – 3y = 8
If we multiply the given equation(1) by 3/2, we get 3x – 3y = 3.
This equation is parallel to 3x – 3y = 8, which means they have the same slope but different y-intercepts.
Parallel lines do not intersect, so this option is incorrect.

C. -10x + 8y = 5
If we multiply the given equation(1) by -5, we get -10x + 10y = -10.
This equation is not parallel to -10x + 8y = 5, which means they have different slopes and intersect at a single point.
This option is correct.

D. -x + y = -1
If we multiply the given equation(1) by -1/2, we get -x + y = -1.
This equation is coincident to -x + y = -1, which means they have the same slope .
Coincident lines do not intersect, so this option is incorrect.

Therefore, the second equation in the system C. -10x + 8y = 5 .

  Question  Hard

One of the two linear equations in a system is −6𝑥 + 7𝑦 = −6. The system has no solution. Which equation could be the second equation in this system?

A. 6𝑥 − 7𝑦 = 0
B. \(-\frac{21x}{4}+\frac{49y}{8}=-\frac{21}{4}\)
C. \(-\frac{21x}{4}-14y=0\)
D. 6𝑥 − 7𝑦 = 6

▶️Answer/Explanation

Ans: A

When a system of linear equations has no solution, it means the lines represented by the equations are parallel, and thus their slopes are equal but their \(y\)-intercepts are not.

The given equation is \(-6 x+7 y=-6\). To find the slope of this line, let’s rewrite it in slope-
\[
\begin{aligned}
& \text { intercept form }(y=m x+b) \text { : } \\
& -6 x+7 y=-6 \\
& 7 y=6 x-6 \\
& y=\frac{6}{7} x-\frac{6}{7}
\end{aligned}
\]

So, the slope of this line is \(m=\frac{6}{7}\).

Option A and D having Equal slope but in D it is coincident line ( infinite no. of solution) as both have -6 intercept.

So correct option is A.

 Question  Hard

−3x + 4y = 4

4x − 3y = 0.5

The solution to the given system of equations is the ordered pair (x,y) What is the value of y ?

▶️Answer/Explanation

2.5

To find the value of \(y\) in the system of equations:
\[
\begin{aligned}
-3x + 4y &= 4 \quad \text{(1)} \\
4x – 3y &= 0.5 \quad \text{(2)}
\end{aligned}
\]

Solve one of the equations for one variable. From equation (2):
\[
4x – 3y = 0.5
\]
\[
4x = 3y + 0.5
\]
\[
x = \frac{3y + 0.5}{4}
\]

Substitute \(x = \frac{3y + 0.5}{4}\) into equation (1):
\[
-3\left(\frac{3y + 0.5}{4}\right) + 4y = 4
\]
\[
-\frac{9y + 1.5}{4} + 4y = 4
\]
\[
-\frac{9y + 1.5}{4} + \frac{16y}{4} = 4
\]
\[
\frac{7y – 1.5}{4} = 4
\]

 Multiply both sides by 4 to clear the fraction:
\[
7y – 1.5 = 16
\]

 Add 1.5 to both sides:
\[
7y = 17.5
\]

Divide by 7:
\[
y = 2.5
\]

  Question   Hard

$
\begin{aligned}
& y=\frac{3}{2} x-\frac{1}{2} \\
& y=\frac{k}{3} x+\frac{1}{3}
\end{aligned}
$

In the system of equations above, \(\mathrm{k}\) is a constant. If the system has no solutions, what is the value of \(\mathrm{k}\) ?

▶️Answer/Explanation

Ans:4.5 Or 9/2

Given the equations:
\[y = \frac{3}{2}x – \frac{1}{2}\]
\[y = \frac{k}{3}x + \frac{1}{3}\]

The slopes of the lines are \(\frac{3}{2}\) and \(\frac{k}{3}\). For the lines to be parallel and have no solutions, these slopes must be equal:
\[\frac{3}{2} = \frac{k}{3}\]

Cross multiply and solve for \(k\):
\[3 \times 3 = 2 \times k\]
\[9 = 2k\]
\[k = \frac{9}{2}\]

So, the value of \(k\) that results in no solutions is \(k = \frac{9}{2}\).

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