Elimination Method Questions in English

(C) Let the two numbers be $x$ and $y,$ where $x > y.$
According to the problem:
$x + y = 20$ (Equation $1$)
$x - y = 8$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (x - y) = 20 + 8$
$2x = 28$
$x = 14$
Substituting $x = 14$ in Equation $1$:
$14 + y = 20$
$y = 20 - 14 = 6$
The two numbers are $14$ and $6.$ The larger number is $14.$

(A) Let $x = \frac{1}{a}$ and $y = \frac{1}{b}$. The equations become:
$8x - 9y = 1$ ---$(1)$
$10x + 6y = 7$ ---$(2)$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$ to eliminate $y$:
$16x - 18y = 2$ ---$(3)$
$30x + 18y = 21$ ---$(4)$
Adding $(3)$ and $(4)$:
$46x = 23 \implies x = \frac{23}{46} = \frac{1}{2}$.
Since $x = \frac{1}{a}$,we have $a = 2$.
Substitute $x = \frac{1}{2}$ into $(1)$:
$8(\frac{1}{2}) - 9y = 1 \implies 4 - 9y = 1 \implies 9y = 3 \implies y = \frac{1}{3}$.
Since $y = \frac{1}{b}$,we have $b = 3$.
Thus,$(a, b) = (2, 3)$.

(C) Given equations are:
$y + x = 2$ --- $(1)$
$y - x = 4$ --- $(2)$
Adding equation $(1)$ and equation $(2)$:
$(y + x) + (y - x) = 2 + 4$
$2y = 6$
$y = 3$
Now,substitute $y = 3$ into equation $(1)$:
$3 + x = 2$
$x = 2 - 3$
$x = -1$
Therefore,the solution is $(x, y) = (-1, 3)$.

(A) Given equations are:
$x + 2y = 5$ --- $(1)$
$3x + 5y = 13$ --- $(2)$
To solve by elimination,multiply equation $(1)$ by $3$:
$3(x + 2y) = 3(5)$
$3x + 6y = 15$ --- $(3)$
Subtract equation $(2)$ from equation $(3)$:
$(3x + 6y) - (3x + 5y) = 15 - 13$
$y = 2$
Substitute $y = 2$ into equation $(1)$:
$x + 2(2) = 5$
$x + 4 = 5$
$x = 5 - 4$
$x = 1$
Thus,the solution is $x = 1, y = 2$.

(B) Given equations are:
$(1)$ $9x - 4y = 14$
$(2)$ $7x - 3y = 11$
To eliminate $x$,multiply equation $(1)$ by $7$ and equation $(2)$ by $9$:
$63x - 28y = 98$ $(3)$
$63x - 27y = 99$ $(4)$
Subtract equation $(3)$ from equation $(4)$:
$(63x - 27y) - (63x - 28y) = 99 - 98$
$63x - 27y - 63x + 28y = 1$
$y = 1$
Thus,the value of $y$ is $1$.

(C) To eliminate the variable $y$ using the elimination method,the coefficients of $y$ in both equations must have the same magnitude but opposite signs.
In equation $(1)$,the coefficient of $y$ is $2$.
In equation $(2)$,the coefficient of $y$ is $-4$.
To make the coefficients of $y$ equal in magnitude,we multiply equation $(1)$ by $2$,which results in $2(x + 2y) = 2(8)$,giving $2x + 4y = 16$.
Now,adding this to equation $(2)$ $(3x - 4y = -6)$ will eliminate $y$.
Therefore,the first equation should be multiplied by $2$.

(A) Given equations are:
$Z+M=88$ --- (Equation $1$)
$Z-M=2$ --- (Equation $2$)
To find the value of $M$,subtract Equation $2$ from Equation $1$:
$(Z+M) - (Z-M) = 88 - 2$
$Z + M - Z + M = 86$
$2M = 86$
$M = \frac{86}{2}$
$M = 43$