Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Oct 27, 2023

Solve the system of linear and quadratic equations $x-6y=10, 3y^2=4x+1$ . Show that $3y^2-24y-41=0$ .

STEP 1

Assumptions1. We have two equations $x -6y =10$ $3y^ =4x +1$ . We need to show that $3y^ -24y -41 =0$

STEP 2

From the first equation, we can express $x$ in terms of $y$ .
$x =6y +10$

STEP 3

Now, we substitute $x$ from the first equation into the second equation.
$3y^2 =(6y +10) +1$

STEP 4

Expand the right side of the equation.
$3y^2 =24y +40 +1$

STEP 5

implify the right side of the equation.
$3y^2 =24y +41$

STEP 6

Rearrange the equation to match the form of the equation we want to prove.
$3y^2 -24y -41 =0$ We have shown that $3y^2 -24y -41 =0$ .

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