Solution of a Linear Equation Questions in English

(D) Given the equation $F = (9/5)C + 32$.
Since $F$ and $C$ are numerically equal,let $F = C = x$.
Substituting $x$ for $F$ and $C$ in the equation:
$x = (9/5)x + 32$
Subtracting $(9/5)x$ from both sides:
$x - (9/5)x = 32$
$(5x - 9x) / 5 = 32$
$-4x / 5 = 32$
$-4x = 32 \times 5$
$-4x = 160$
$x = 160 / -4$
$x = -40$
Therefore,$F$ and $C$ are numerically equal when $F = -40$.

(A) To express the equation $5x + 2y = 10$ in the $y$-form,we isolate $y$ on one side of the equation.
First,subtract $5x$ from both sides:
$2y = 10 - 5x$
Next,divide both sides by $2$:
$y = \frac{10 - 5x}{2}$

(B) Given the equation is $ax + 3y = 21$.
Since the graph passes through the point $(5, 2)$,the coordinates $x = 5$ and $y = 2$ must satisfy the equation.
Substituting the values of $x$ and $y$ into the equation:
$a(5) + 3(2) = 21$
$5a + 6 = 21$
$5a = 21 - 6$
$5a = 15$
$a = \frac{15}{5}$
$a = 3$
Therefore,the value of $a$ is $3$.

(C) Given the equation: $3y - 2x - 60 = 0$
To express it in the $y$-form,we isolate $y$ on one side of the equation.
First,add $2x$ and $60$ to both sides:
$3y = 2x + 60$
Next,divide both sides by $3$:
$y = \frac{2x + 60}{3}$
Thus,the correct option is $C$.

(D) To determine if a point $(x, y)$ lies on the graph of the equation $3x + 2y = 18$,we substitute the $x$ and $y$ values into the equation and check if the left-hand side equals the right-hand side $(18)$.
For option $A$ $(6, 0)$: $3(6) + 2(0) = 18 + 0 = 18$. (Point lies on the graph)
For option $B$ $(0, 9)$: $3(0) + 2(9) = 0 + 18 = 18$. (Point lies on the graph)
For option $C$ $(2, 6)$: $3(2) + 2(6) = 6 + 12 = 18$. (Point lies on the graph)
For option $D$ $(3, 2)$: $3(3) + 2(2) = 9 + 4 = 13$. Since $13 \neq 18$,the point $(3, 2)$ does not lie on the graph.

(A) To find the point where the graph of the equation $2x + y = 10$ intersects the $x$-axis,we set the $y$-coordinate to $0$.
Substituting $y = 0$ into the equation: $2x + 0 = 10$.
This simplifies to $2x = 10$.
Dividing both sides by $2$,we get $x = 5$.
Therefore,the point of intersection is $(5, 0)$.

(C) To determine which point lies on the graph of the equation $3x - 5y = 15$,we substitute the coordinates $(x, y)$ of each option into the equation:
$A) (0, 0): 3(0) - 5(0) = 0 \neq 15$
$B) (-5, 0): 3(-5) - 5(0) = -15 \neq 15$
$C) (10, 3): 3(10) - 5(3) = 30 - 15 = 15$. Since the left-hand side equals the right-hand side,this point lies on the graph.
$D) (0, 3): 3(0) - 5(3) = -15 \neq 15$
Therefore,the correct point is $(10, 3)$.

(A) To express the equation $3x - 4y = 24$ in $y$-form,we need to isolate $y$ on one side of the equation.
Step $1$: Subtract $3x$ from both sides: $-4y = -3x + 24$.
Step $2$: Divide both sides by $-4$: $y = \frac{-3x + 24}{-4}$.
Step $3$: Simplify the expression: $y = \frac{-3x}{-4} + \frac{24}{-4}$.
Step $4$: Resulting in $y = \frac{3}{4}x - 6$.

(A) Given the equation: $3x + 2y = 18$.
Substitute $x = 5$ into the equation:
$3(5) + 2y = 18$
$15 + 2y = 18$
Subtract $15$ from both sides:
$2y = 18 - 15$
$2y = 3$
Divide by $2$:
$y = \frac{3}{2} = 1.5$.

(B) Given that the point $(5, 2)$ lies on the graph of the equation $3x + 2y = k$.
This means that the coordinates $x = 5$ and $y = 2$ must satisfy the equation.
Substituting the values of $x$ and $y$ into the equation:
$3(5) + 2(2) = k$
$15 + 4 = k$
$k = 19$
Therefore,the value of $k$ is $19$.

(C) The general form of a linear equation in two variables is $ax + by + c = 0$.
If $c = 0$,the equation becomes $ax + by = 0$.
For any linear equation of the form $ax + by = 0$,if we substitute $x = 0$ and $y = 0$,we get $a(0) + b(0) = 0$,which is $0 = 0$.
Since the point $(0, 0)$ satisfies the equation,the line must pass through the origin $(0, 0)$.

(A) Given the linear equation $5x - ay = 7$.
Since $(3, 2)$ is a solution,we substitute $x = 3$ and $y = 2$ into the equation.
$5(3) - a(2) = 7$
$15 - 2a = 7$
$-2a = 7 - 15$
$-2a = -8$
$a = \frac{-8}{-2}$
$a = 4$
Therefore,the correct option is $A$.

(A) The equation $y=5$ represents a line where the $y$-coordinate is always $5$ for any value of $x$.
Since the $y$-coordinate remains constant,the line does not change its vertical position,meaning it is a horizontal line.
$A$ horizontal line is always parallel to the $x$-axis.
Therefore,the graph of $y=5$ is parallel to the $x$-axis.

(C) Given the linear equation: $5x + 3y = 90$.
Substitute the value $x = 12$ into the equation:
$5(12) + 3y = 90$
$60 + 3y = 90$
Subtract $60$ from both sides:
$3y = 90 - 60$
$3y = 30$
Divide by $3$:
$y = 10$.

(D) Given that $(-2, -3)$ is a solution of the equation $ax - 5y = 21$.
This means that $x = -2$ and $y = -3$ satisfy the equation.
Substituting these values into the equation:
$a(-2) - 5(-3) = 21$
$-2a + 15 = 21$
$-2a = 21 - 15$
$-2a = 6$
$a = 6 / -2$
$a = -3$

(A) To find the point where the graph of the linear equation $5x - 3y = 20$ intersects the $x$-axis,we set the $y$-coordinate to $0$.
Substituting $y = 0$ into the equation:
$5x - 3(0) = 20$
$5x = 20$
$x = 20 / 5$
$x = 4$
Therefore,the graph intersects the $x$-axis at the point $(4, 0)$.

(C) linear equation in two variables is of the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
For any value assigned to one variable (e.g.,$x$),we can find a corresponding value for the other variable (e.g.,$y$).
Since there are infinitely many real numbers that can be assigned to $x$,there are infinitely many corresponding values for $y$.
Therefore,a linear equation in two variables has infinitely many solutions.

(D) The standard form of a linear equation in two variables is $ax + by + c = 0$.
Given equation: $2x + 3y = 12$.
Subtracting $12$ from both sides,we get: $2x + 3y - 12 = 0$.
Comparing this with $ax + by + c = 0$,we identify the coefficients as:
$a = 2$
$b = 3$
$c = -12$
Now,calculate the sum $a + b + c$:
$a + b + c = 2 + 3 + (-12)$
$a + b + c = 5 - 12 = -7$.
Therefore,the value of $a + b + c$ is $-7$.

(B) The given equation is $x+y=0$,which can be rewritten as $y = -x$.
For $x = 0$,$y = 0$,so the line passes through the origin $(0, 0)$.
If $x > 0$,then $y < 0$,which corresponds to the fourth quadrant.
If $x < 0$,then $y > 0$,which corresponds to the second quadrant.
Therefore,the graph passes through the second and fourth quadrants.

(B) Given the linear equation $3x - 4y = k$.
Since $(2, 5)$ is a solution to this equation,it must satisfy the equation.
Substitute $x = 2$ and $y = 5$ into the equation:
$3(2) - 4(5) = k$
$6 - 20 = k$
$k = -14$
Therefore,the value of $k$ is $-14$.

(D) Given that $(5, 2)$ is a solution of the equation $3x + ky = 25$.
This means that when $x = 5$ and $y = 2$,the equation must be satisfied.
Substituting the values of $x$ and $y$ into the equation:
$3(5) + k(2) = 25$
$15 + 2k = 25$
Subtracting $15$ from both sides:
$2k = 25 - 15$
$2k = 10$
Dividing by $2$:
$k = 5$
Therefore,the value of $k$ is $5$.

(B) Given the equation: $0x + 3y = 21$.
Simplifying this,we get $3y = 21$,which implies $y = 7$.
The equation $y = 7$ represents a horizontal line parallel to the $x$-axis.
$A$ line that is parallel to the $x$-axis is perpendicular to the $y$-axis.
Therefore,the graph of the equation $0x + 3y = 21$ is perpendicular to the $y$-axis.

(B) To find the common solution,we solve the system of linear equations:
$1$) $x + y = 0$
$2$) $x - y = 0$
Adding equation $(1)$ and $(2)$:
$(x + y) + (x - y) = 0 + 0$
$2x = 0$
$x = 0$
Substituting $x = 0$ into equation $(1)$:
$0 + y = 0$
$y = 0$
Therefore,the common solution is $x = 0, y = 0$.

(C) Given the equation is $3x - 4y = k$.
Since $(4, 3)$ is a solution to this equation,we substitute $x = 4$ and $y = 3$ into the equation.
$3(4) - 4(3) = k$
$12 - 12 = k$
$0 = k$
Therefore,the value of $k$ is $0$.

(A) To express the equation $y = x + 5$ in the form $ax + by + c = 0$,we rearrange the terms:
$1$. Subtract $x$ and $5$ from both sides: $-x + y - 5 = 0$.
$2$. Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = -1$
$b = 1$
$c = -5$.