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Maths 3 Marks Question

P-1 Linear Equations in Two Variables · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 109 questions.

Questions · Page 3 of 3

3 Marks Question

Question 1013 Marks
Solve : 15x + 17y = 21; 17x + 15y = 11
Answer
$\begin{gathered}
15 x+17 y=21 \ldots \text { (I) } \\
17 x+15 y=11 \ldots \text { (II) }
\end{gathered}$
In the two equations above, the coefficients of $x$ and $y$ are interchanged. While solving such equations we get two simple equations by adding and subtracting the given equations. After solving these equations, we can easily find the solution.
Let's add the two given equations.
$\begin{aligned}
15 x+17 y & =21 \\
+17 x+15 y & =11 \\
\hline 32 x+32 y & =32
\end{aligned}$
Dividing both sides of the equation by 32 .
$x+y=1 \ldots \text {(III)}$
Now, let's subtract equation (II) from (I)
$\begin{aligned}
& \quad 15 x+17 y=21 \\
& \begin{array}{r}
+ \_17 x \pm 15 y=\_11 \\
\hline-2 x+2 y=10
\end{array} \\
\end{aligned}$
dividing the equation by 2 .
$-x+y=5 \ldots \text { (IV) }$
Now let's add equations (III) and (V).
$\begin{aligned}
& \quad x+y=1 \\
+ & -x+y=5 \\
& \hline \therefore \quad 2 y=6 \\
& \therefore \quad y=3 \\
\end{aligned}$
Place this value in equation (III).
$\begin{array}{ll} 
x+y=1 \\
\therefore x+3=1 \\
\therefore x=1-3 \quad \therefore x=-2 \\
(x, y)=(-2,3)  \text { is the solution. }
\end{array}$
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Question 1023 Marks
Solve :
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
Answer
$\begin{aligned}
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
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Question 1033 Marks
Sum of two number is 45 and the greater number is twice the smaller number. Find the numbers.
Answer
30,15
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Question 1043 Marks
Shabana's age 10 years hence, will be twice juhi's present age. 6 years back shabana's age was $\frac{5}{3}$ times Juhi's at that time find their present ages.
Answer
26 years, 18 years
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Question 1053 Marks
If 1 is added to the numerator of a certain fraction its value becomes $\frac{1}{2}$ and if 1 is added to its denometer $\frac{1}{3}$. Find the original fraction.
Answer
$\frac{3}{8}$
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3 Marks Question - Page 3 - Maths STD 10 Questions - Vidyadip