Practice Questions
Solve: $3x + 5 = 20$. What is the value of $x$?
$3x = 20 - 5 = 15$. $x = 15/3 = 5$.
If $2x + 3y = 12$ and $x - y = 1$, find $x$.
From second: $x = y + 1$. Substitute: $2(y+1) + 3y = 12 \Rightarrow 5y = 10 \Rightarrow y = 2$. So $x = 3$.
The sum of two numbers is 25 and their difference is 5. Find the larger number.
$x + y = 25$ and $x - y = 5$. Adding: $2x = 30 \Rightarrow x = 15$.
If $4x - 3y = 5$ and $2x + y = 5$, what is the value of $y$?
From eq2: $y = 5 - 2x$. Substitute into eq1: $4x - 3(5-2x) = 5 \Rightarrow 10x = 20 \Rightarrow x = 2$. Then $y = 5 - 2(2) = 1$.
Solve: $4x + 7 = 3x + 15$.
$4x - 3x = 15 - 7 \Rightarrow x = 8$.
Solve the system: $3x + 2y = 16$, $x - y = 1$.
$x = y + 1$. $3(y+1) + 2y = 16 \Rightarrow 5y = 13 \Rightarrow y = 2.6$, $x = 3.6$.
The sum of two numbers is 30 and their difference is 4. Find the numbers.
$x + y = 30$, $x - y = 4$. Adding: $2x = 34 \Rightarrow x = 17$, $y = 13$.
A father is three times as old as his son. After 10 years, he will be twice as old. Find son's present age.
Let son = $x$, father = $3x$. After 10 yrs: $3x + 10 = 2(x + 10) \Rightarrow 3x + 10 = 2x + 20 \Rightarrow x = 10$.
The cost of 3 pens and 2 notebooks is ₹94. The cost of 2 pens and 3 notebooks is ₹91. Find cost of a pen.
$3p + 2n = 94$, $2p + 3n = 91$. Multiply eq1 by 3 and eq2 by 2: $9p + 6n = 282$, $4p + 6n = 182$. Subtract: $5p = 100 \Rightarrow p = 20$.
Solve using substitution: $5x - 2y = 4$, $3x + y = 11$.
From eq2: $y = 11 - 3x$. Substitute: $5x - 2(11-3x) = 4 \Rightarrow 11x = 26 \Rightarrow x = 2$, $y = 5$.
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