the system of equations below has no solution. which equation could represent a linear combination of the…

the system of equations below has no solution. which equation could represent a linear combination of the system?\n$$\n\\left\\{\\begin{array}{l}\n\\frac{2}{3} x+\\frac{5}{2} y=15 \\\\\n4 x+15 y=12\n\\end{array}\\right.\n$$\n$$\n\\frac{4}{3} x=42\n$$\n$$\n0=-78\n$$\n$$\n\\frac{15}{2} y=33\n$$\n$$\n0=0\n$$

the system of equations below has no solution. which equation could represent a linear combination of the system?\n$$\n\\left\\{\\begin{array}{l}\n\\frac{2}{3} x+\\frac{5}{2} y=15 \\\\\n4 x+15 y=12\n\\end{array}\\right.\n$$\n$$\n\\frac{4}{3} x=42\n$$\n$$\n0=-78\n$$\n$$\n\\frac{15}{2} y=33\n$$\n$$\n0=0\n$$

Answer

Explanation:

Step1: Multiply the first equation by 6

Multiply (\frac{2}{3}x+\frac{5}{2}y = 15) by 6. Using the distributive property (a(b + c)=ab+ac), we get (6\times\frac{2}{3}x+6\times\frac{5}{2}y=6\times15), which simplifies to (4x + 15y=90).

Step2: Subtract the second equation from the new - formed equation

We have the system: (\begin{cases}4x + 15y=90\4x + 15y=12\end{cases}) Subtract the second equation (4x + 15y = 12) from the first equation (4x+15y = 90). ((4x + 15y)-(4x + 15y)=90 - 12). The left - hand side: (4x+15y-4x - 15y=(4x-4x)+(15y - 15y)=0). The right - hand side: (90-12 = 78). So, (0=-78) (since (90-12) gives a non - zero result when the left - hand side is (0)).

Answer:

(0=-78)