Question
Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution $:2( x - 3 ) + 3( y - 5 ) = 0,5( x - 1 ) + 4( y - 4 ) = 0$
✓
Answer
Given equations are
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3),$ we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3),$ we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.
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
The base of a triangle field and its height are in the ratio $2:5.$ If the cost of cultivating the field at Rs.42 per $sq. m$ is $Rs.7560$, find its base and height.
→Evaluate :$\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times\left(\frac{49}{9}\right)^{\frac{3}{2}} \div\left(\frac{343}{216}\right)^{\frac{2}{3}}$
→In the adjoining figure, in $\triangle ABC , AD$ is the median through A and E is the mid-point of AD . If BE produced meets AC in F , prove that $AF =\frac{1}{3} AC$.
→If $\left(x+\frac{1}{x}\right)=5$, find the values of $\left(x^4+\frac{1}{x^4}\right)$.
→The ages of $A$ and $B$ are in the ratio $7:5.$ Ten years hence, the ratio of their ages will be $9:7.$ Find their ages.
→$\text{ABCD}$ is a quadrilateral in which diagonals $AC$ and $BD$ intersect at a point $O$. Prove that: area $\triangle AOD +$ area $\triangle BOC +$ area $\triangle ABO +$ area $\triangle CDO.$
→In the given figure, arc $AB =$ twice arc BC and $\angle AOB =80^{\circ}$. Find:
(i) $\angle BOC$
(ii) $\angle OAC$.
→(i) $\angle BOC$
(ii) $\angle OAC$.
A ladder $25\ m$ long reaches a window of a building $20\ m$ above the ground. Determine the distance of the foot of the ladder from the building.
→If $\frac{2+\sqrt{ } 5}{2-\sqrt{ } 5}=x$ and $\frac{2-\sqrt{ } 5}{2+\sqrt{ } 5}=y ;$ find the value of $x^2-y^2$
→
AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take $\pi$ = 3.14)
[Hint: Perimeter of the shaded parts = 2(OA + OD + arc AD)
$\begin{aligned} & =2\left(8+8+\frac{3.14 \times 8 \times 90}{360}\right) cm =57.12 cm \\ \text { Area of the shaded parts } & =2 \times(\text { area of quadrant } O A D) \\ & \left.=\frac{(2 \times \pi \times 8 \times 8 \times 90)}{360} cm^2=100.48 cm^2\right]\end{aligned}$