Solve the following pair of linear equations | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$.

Vedclass · Accepted Answer

$(2, -3)$

Vedclass · Answer

(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$. If $y = 0$,then $x = -4$. The line passes through $(0, -2)$ and $(-4, 0)$.$2$. For the second equation $3x + 4y = -6$,if $x = 0$,then $y = -1.5$. If $y = 0$,then $x = -2$. The line passes through $(0, -1.5)$ and $(-2, 0)$.$3$. Solving the equations algebraically: Multiply the first equation by $2$: $2x + 4y = -8$. Subtract this from the second equation: $(3x + 4y) - (2x + 4y) = -6 - (-8)$,which gives $x = 2$.$4$. Substitute $x = 2$ into $x + 2y = -4$: $2 + 2y = -4 \implies 2y = -6 \implies y = -3$.$5$. The intersection point is $(2, -3)$.

Solve the following pair of linear equations in two variables using a graph: $x + 2y = -4$ and $3x + 4y = -6$.

Similar Questions

Solve the following pairs of equations:
$\frac{1}{2x} - \frac{1}{y} = -1$
$\frac{1}{x} + \frac{1}{2y} = 8, \quad x, y \neq 0$

Obtain the solution of the following pair of equations by the cross-multiplication method: $ax + by = 1$ and $bx + ay = \frac{2ab}{a^2 + b^2}$.

The pair of equations $5x - y = 9$ and $10x - 18 = 2y$ is $\ldots \ldots \ldots \ldots$

If the sum of two numbers is $15$ and the subtraction is $1,$ then the smaller number of these two numbers is.........

Solve the following pair of linear equations by the method of elimination: $0.3 x + 0.4 y = 1$ and $6 x + 8 y = 20$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module