Question
Solve: $\mathrm{y}+\frac{\mathrm{y}}{2}=\frac{7}{4}-\frac{\mathrm{y}}{4}$
✓
Answer
$\begin{aligned} & y+\frac{y}{2}=\frac{7}{4}-\frac{y}{4} \\ & \Rightarrow \frac{y}{1}+\frac{y}{2}=\frac{7}{4}-\frac{y}{4} \\ & \Rightarrow \frac{4 y+2 y=7-y}{4} \\ & \Rightarrow 4 y+2 y=7-y \\ & \Rightarrow 4 y+2 y \Rightarrow 7 y=7 \\ & \Rightarrow y=\frac{7}{7}=1 \\ & \therefore y=1\end{aligned}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
What number: When decreased by 50% becomes 50?
Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figure:
Add the rational number : $\frac{21}{-4}$ and $\frac{-11}{8}$
→The area of a right-angled triangle is $160 cm^2$. If its one leg is $16$ cm long, find the length of the other leg.
→An article bought for Rs. 350 is sold at a profit of 20%. Find its selling price.
Express : 2.8 mm as a per cent of 2.24 cm
Show that 2, 12 and 72 are in continued proportion.
Simplify and express the Solution in the positive exponent form:
$\frac{\left(2^3\right)^5 \times 5^4}{4^3 \times 5^2}$
→$\frac{\left(2^3\right)^5 \times 5^4}{4^3 \times 5^2}$
Subtract : $5 x^2-3 x y-7 y^2$ from $3 x^2-x y-2 y^2$
→Given below are the lengths of the legs of right $\triangle A B C$. In each case, find the length of the hypotenuse of $\triangle A B C$ :
(i) $a=32 cm, b=60 cm$
(ii) $a=28 m, b=45 m$
(iii) $a=32 cm, b=24 cm$
→(i) $a=32 cm, b=60 cm$
(ii) $a=28 m, b=45 m$
(iii) $a=32 cm, b=24 cm$